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<?xml version="1.0" encoding="utf-8"?>
<feed xmlns="http://www.w3.org/2005/Atom">
<title>BlueSky</title>
<link href="http://zuqis.top/atom.xml" rel="self"/>
<link href="http://zuqis.top/"/>
<updated>2023-11-16T15:15:46.134Z</updated>
<id>http://zuqis.top/</id>
<author>
<name>ListMatrix</name>
</author>
<generator uri="https://hexo.io/">Hexo</generator>
<entry>
<title>洛谷P1359租用游艇</title>
<link href="http://zuqis.top/2023/11/16/%E6%B8%B8%E8%89%87%E9%97%AE%E9%A2%98/"/>
<id>http://zuqis.top/2023/11/16/%E6%B8%B8%E8%89%87%E9%97%AE%E9%A2%98/</id>
<published>2023-11-16T14:48:15.000Z</published>
<updated>2023-11-16T15:15:46.134Z</updated>
<content type="html"><![CDATA[<h2 id="题目描述"><a href="#题目描述" class="headerlink" title="题目描述"></a>题目描述</h2><p>长江游艇俱乐部在长江上设置了 $n$ 个游艇出租站 $1,2,\cdots,n$。游客可在这些游艇出租站租用游艇，并在下游的任何一个游艇出租站归还游艇。游艇出租站 $i$ 到游艇出租站 $j$ 之间的租金为 $r(i,j)$（$1\le i\lt j\le n$）。试设计一个算法，计算出从游艇出租站 $1$ 到游艇出租站 $n$ 所需的最少租金。</p><h2 id="输入格式"><a href="#输入格式" class="headerlink" title="输入格式"></a>输入格式</h2><p>第一行中有一个正整数 $n$，表示有 $n$ 个游艇出租站。接下来的 $n-1$ 行是一个半矩阵 $r(i,j)$（$1\le i<j\le n$）。</p><h2 id="输出格式"><a href="#输出格式" class="headerlink" title="输出格式"></a>输出格式</h2><p>输出计算出的从游艇出租站 $1$ 到游艇出租站 $n$ 所需的最少租金。</p><h2 id="样例-1"><a href="#样例-1" class="headerlink" title="样例 #1"></a>样例 #1</h2><h3 id="样例输入-1"><a href="#样例输入-1" class="headerlink" title="样例输入 #1"></a>样例输入 #1</h3><figure class="highlight plaintext"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br></pre></td><td class="code"><pre><span class="line">3</span><br><span class="line">5 15</span><br><span class="line">7</span><br></pre></td></tr></table></figure><h3 id="样例输出-1"><a href="#样例输出-1" class="headerlink" title="样例输出 #1"></a>样例输出 #1</h3><figure class="highlight plaintext"><table><tr><td class="gutter"><pre><span class="line">1</span><br></pre></td><td class="code"><pre><span class="line">12</span><br></pre></td></tr></table></figure><h2 id="提示"><a href="#提示" class="headerlink" title="提示"></a>提示</h2><h2 id="n-le-200-，保证计算过程中任何时刻数值都不超过-10-6-。"><a href="#n-le-200-，保证计算过程中任何时刻数值都不超过-10-6-。" class="headerlink" title="$n\le 200$，保证计算过程中任何时刻数值都不超过 $10^6$。"></a>$n\le 200$，保证计算过程中任何时刻数值都不超过 $10^6$。</h2><p>##直接dp<br> 状态转移方程：<code>dp[j] = min(dp[j], f[i][j] + dp[i])</code>，只需要从1到n一遍一遍的更新dp。（注意dp要初始化一个很大的值）<br>###代码</p><figure class="highlight plaintext"><figcaption><span><iostream></span></figcaption><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br></pre></td><td class="code"><pre><span class="line">#include <algorithm></span><br><span class="line">#include <ctime></span><br><span class="line">using namespace std;</span><br><span class="line">typedef long long ll;</span><br><span class="line"></span><br><span class="line">int n, dp[205], f[200][200];</span><br><span class="line">int main()</span><br><span class="line">{</span><br><span class="line">#ifdef LOCALFLAG</span><br><span class="line"> freopen(".in", "rt", stdin);</span><br><span class="line"> freopen(".out", "wt", stdout);</span><br><span class="line"> auto t1 = clock();</span><br><span class="line">#endif</span><br><span class="line"></span><br><span class="line"> cin >> n;</span><br><span class="line"> for (int i = 1; i <= n;i++){</span><br><span class="line"> for (int j = i + 1; j <= n;j++)</span><br><span class="line"> cin >> f[i][j];</span><br><span class="line"> dp[i] = 1e9; //初始化dp为无穷大</span><br><span class="line"> }</span><br><span class="line"> dp[1] = f[1][1];</span><br><span class="line"> for (int i = 1; i <= n;i++)</span><br><span class="line"> for (int j = i + 1; j <= n;j++)</span><br><span class="line"> dp[j] = min(dp[j], f[i][j] + dp[i]);</span><br><span class="line"> cout << dp[n]<<endl;</span><br><span class="line"></span><br><span class="line">#ifdef LOCALFLAG</span><br><span class="line"> auto t2 = clock();</span><br><span class="line"> cout << "Time cost= " << t2 - t1 << endl;</span><br><span class="line">#endif</span><br><span class="line"> return 0;</span><br><span class="line">}</span><br></pre></td></tr></table></figure><p> 初次之外还有floyd法和dijkstra法，下面是代码：</p><p>##Floyd法</p><figure class="highlight plaintext"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br></pre></td><td class="code"><pre><span class="line">#include <iostream></span><br><span class="line">#include <algorithm></span><br><span class="line">#include <ctime></span><br><span class="line">#include <string.h></span><br><span class="line">using namespace std;</span><br><span class="line">typedef long long ll;</span><br><span class="line"></span><br><span class="line">int n,x, f[200][200];</span><br><span class="line">int edge(int u,int v,int x){ //建立边的函数</span><br><span class="line"> f[u][v] = x;</span><br><span class="line">}</span><br><span class="line">int main()</span><br><span class="line">{</span><br><span class="line">#ifdef LOCALFLAG</span><br><span class="line"> freopen(".in", "rt", stdin);</span><br><span class="line"> freopen(".out", "wt", stdout);</span><br><span class="line"> auto t1 = clock();</span><br><span class="line">#endif</span><br><span class="line"></span><br><span class="line"> cin >> n;</span><br><span class="line"> memset(f, 0x3f, sizeof(f)); //初始化数组</span><br><span class="line"> for (int i = 1; i <= n;i++){</span><br><span class="line"> for (int j = i + 1; j <= n;j++){</span><br><span class="line"> if(i == j)</span><br><span class="line"> f[i][j] = 0;</span><br><span class="line"> else{</span><br><span class="line"> cin >> x;</span><br><span class="line"> edge(i, j, x);</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> //floyd</span><br><span class="line"> for (int k = 1; k <= n; k++) // floyd</span><br><span class="line"> for (int i = 1; i <= n; i++)</span><br><span class="line"> for (int j = 1; j <= n; j++)</span><br><span class="line"> {</span><br><span class="line"> f[i][j] = min(f[i][j], f[i][k] + f[k][j]);</span><br><span class="line"> }</span><br><span class="line"> cout << f[1][n] << endl;</span><br><span class="line"> </span><br><span class="line">#ifdef LOCALFLAG</span><br><span class="line"> auto t2 = clock();</span><br><span class="line"> cout << "Time cost= " << t2 - t1 << endl;</span><br><span class="line">#endif</span><br><span class="line"> return 0;</span><br><span class="line">}</span><br></pre></td></tr></table></figure><p>##dijkstra法</p><figure class="highlight plaintext"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br><span class="line">47</span><br><span class="line">48</span><br><span class="line">49</span><br></pre></td><td class="code"><pre><span class="line">#include <iostream></span><br><span class="line">#include <cstring></span><br><span class="line">#include <algorithm></span><br><span class="line">using namespace std;</span><br><span class="line">const int INF = 9999999;</span><br><span class="line">int main()</span><br><span class="line">{</span><br><span class="line"> int n,g[205][205];</span><br><span class="line"> cin >> n;</span><br><span class="line"> for(int i = 1;i <= n;i++)</span><br><span class="line"> for(int j = 1;j <= n;j++)</span><br><span class="line"> {</span><br><span class="line"> if(i == j)</span><br><span class="line"> g[i][j] = 0;</span><br><span class="line"> else</span><br><span class="line"> g[i][j] = INF;</span><br><span class="line"> }</span><br><span class="line"> for (int i = 1;i <= n-1;i++)</span><br><span class="line"> for(int j = i+1;j <= n;j++)</span><br><span class="line"> {</span><br><span class="line"> cin >> g[i][j];</span><br><span class="line"> }</span><br><span class="line"> //dijstkra</span><br><span class="line"> int dis[205],book[205];</span><br><span class="line"> memset(book,0,sizeof(book));</span><br><span class="line"> for(int i = 1;i <= n;i++)</span><br><span class="line"> dis[i] = g[1][i];</span><br><span class="line"> dis[1] = 0;</span><br><span class="line"> book[1] = 1;</span><br><span class="line"> for(int i = 2;i <= n;i++)</span><br><span class="line"> {</span><br><span class="line"> int mind = INF,u;</span><br><span class="line"> for(int j = 1;j <= n;j++)</span><br><span class="line"> {</span><br><span class="line"> if(!book[j] && dis[j] < mind)</span><br><span class="line"> {</span><br><span class="line"> mind = dis[j];</span><br><span class="line"> u = j;</span><br><span class="line"> } </span><br><span class="line"> }</span><br><span class="line"> book[u] = 1;</span><br><span class="line"> for(int j = 1;j <= n;j++)</span><br><span class="line"> {</span><br><span class="line"> dis[j] = min(dis[j],dis[u]+g[u][j]);</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> cout << dis[n];</span><br><span class="line"> return 0;</span><br><span class="line">}</span><br></pre></td></tr></table></figure>]]></content>
<summary type="html"><h2 id="题目描述"><a href="#题目描述" class="headerlink" title="题目描述"></a>题目描述</h2><p>长江游艇俱乐部在长江上设置了 $n$ 个游艇出租站 $1,2,\cdots,n$。游客可在这些游艇出租站租用游艇，并在下游的</summary>
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<category term="最短路径" scheme="http://zuqis.top/tags/%E6%9C%80%E7%9F%AD%E8%B7%AF%E5%BE%84/"/>
</entry>
<entry>
<title>层次分析法</title>
<link href="http://zuqis.top/2023/11/16/%E5%B1%82%E6%AC%A1%E5%88%86%E6%9E%90%E6%B3%95/"/>
<id>http://zuqis.top/2023/11/16/%E5%B1%82%E6%AC%A1%E5%88%86%E6%9E%90%E6%B3%95/</id>
<published>2023-11-16T11:12:29.000Z</published>
<updated>2023-11-16T13:46:37.033Z</updated>
<category term="数学建模" scheme="http://zuqis.top/tags/%E6%95%B0%E5%AD%A6%E5%BB%BA%E6%A8%A1/"/>
</entry>
<entry>
<title>洛谷P1466[USACO2.2] 集合 Subset Sums</title>
<link href="http://zuqis.top/2023/11/14/p1466%E9%9B%86%E5%90%88/"/>
<id>http://zuqis.top/2023/11/14/p1466%E9%9B%86%E5%90%88/</id>
<published>2023-11-14T08:51:34.885Z</published>
<updated>2023-11-14T23:51:53.475Z</updated>
<content type="html"><![CDATA[<h2 id="题目描述"><a href="#题目描述" class="headerlink" title="题目描述"></a>题目描述</h2><p>对于从 $1\sim n$ 的连续整数集合，能划分成两个子集合，且保证每个集合的数字和是相等的。举个例子，如果 $n=3$，对于 ${1,2,3}$ 能划分成两个子集合，每个子集合的所有数字和是相等的：</p><p>${3}$ 和 ${1,2}$ 是唯一一种分法（交换集合位置被认为是同一种划分方案，因此不会增加划分方案总数）<br>如果 $n=7$，有四种方法能划分集合 ${1,2,3,4,5,6,7 }$，每一种分法的子集合各数字和是相等的:</p><p>${1,6,7}$ 和 ${2,3,4,5}$<br>${2,5,7}$ 和 ${1,3,4,6}$<br>${3,4,7}$ 和 ${1,2,5,6}$<br>${1,2,4,7}$ 和 ${3,5,6}$ </p><p>给出 $n$，你的程序应该输出划分方案总数。</p><h2 id="输入格式"><a href="#输入格式" class="headerlink" title="输入格式"></a>输入格式</h2><p>输入文件只有一行，且只有一个整数 $n$</p><h2 id="输出格式"><a href="#输出格式" class="headerlink" title="输出格式"></a>输出格式</h2><p>输出划分方案总数。</p><h2 id="样例-1"><a href="#样例-1" class="headerlink" title="样例 #1"></a>样例 #1</h2><h3 id="样例输入-1"><a href="#样例输入-1" class="headerlink" title="样例输入 #1"></a>样例输入 #1</h3><figure class="highlight plaintext"><table><tr><td class="gutter"><pre><span class="line">1</span><br></pre></td><td class="code"><pre><span class="line">7</span><br></pre></td></tr></table></figure><h3 id="样例输出-1"><a href="#样例输出-1" class="headerlink" title="样例输出 #1"></a>样例输出 #1</h3><figure class="highlight plaintext"><table><tr><td class="gutter"><pre><span class="line">1</span><br></pre></td><td class="code"><pre><span class="line">4</span><br></pre></td></tr></table></figure><h2 id="提示"><a href="#提示" class="headerlink" title="提示"></a>提示</h2><p>【数据范围】<br>对于 $100%$ 的数据，$1\le n \le 39$。</p><p>翻译来自NOCOW</p><p>USACO 2.2</p><hr><p>##思路</p><p>分为两个集合，那就将集合看成一个背包，在N个数中挑选N/2个数使这个背包达到最大容量<code>sum = n*(n+1)/4</code>,最后求出的<code>dp[sum]</code>是所有方案的总和（选的方案和不选的方案），将总数除以2便得到了答案。</p><p>##代码</p><figure class="highlight plaintext"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br></pre></td><td class="code"><pre><span class="line">#include <iostream></span><br><span class="line">#include <algorithm></span><br><span class="line">#include <ctime></span><br><span class="line">using namespace std;</span><br><span class="line">typedef long long ll;</span><br><span class="line">ll n, sum, dp[810];</span><br><span class="line">int main()</span><br><span class="line">{</span><br><span class="line">#ifdef LOCALFLAG</span><br><span class="line"> freopen(".in", "rt", stdin);</span><br><span class="line"> freopen(".out", "wt", stdout);</span><br><span class="line"> auto t1 = clock();</span><br><span class="line">#endif</span><br><span class="line"></span><br><span class="line"></span><br><span class="line"> cin >> n;</span><br><span class="line"> sum = n * (n + 1) / 2; </span><br><span class="line"> if(sum%2 == 1){</span><br><span class="line"> cout << "0" << endl;</span><br><span class="line"> return 0;</span><br><span class="line"> }</span><br><span class="line"> sum /= 2;</span><br><span class="line"> dp[0] = 1;</span><br><span class="line"> for (int i = 1; i <= n;i++)</span><br><span class="line"> for (int j = sum; j >= i;j--)</span><br><span class="line"> dp[j] += dp[j - i];</span><br><span class="line"> cout << dp[sum] / 2 << endl;</span><br><span class="line"></span><br><span class="line">#ifdef LOCALFLAG</span><br><span class="line"> auto t2 = clock();</span><br><span class="line"> cout << "Time cost= " << t2 - t1 << endl;</span><br><span class="line">#endif</span><br><span class="line"> return 0;</span><br><span class="line">}</span><br></pre></td></tr></table></figure>]]></content>
<summary type="html"><h2 id="题目描述"><a href="#题目描述" class="headerlink" title="题目描述"></a>题目描述</h2><p>对于从 $1\sim n$ 的连续整数集合，能划分成两个子集合，且保证每个集合的数字和是相等的。举个例子，如果 $n&#x3</summary>
</entry>
<entry>
<title>洛谷P2871[USACO07DEC] Charm Bracelet S</title>
<link href="http://zuqis.top/2023/11/14/01%E8%83%8C%E5%8C%85/"/>
<id>http://zuqis.top/2023/11/14/01%E8%83%8C%E5%8C%85/</id>
<published>2023-11-14T07:28:19.459Z</published>
<updated>2023-11-15T00:09:21.967Z</updated>
<content type="html"><![CDATA[<h2 id="题目描述"><a href="#题目描述" class="headerlink" title="题目描述"></a>题目描述</h2><p>Bessie has gone to the mall’s jewelry store and spies a charm bracelet. Of course, she’d like to fill it with the best charms possible from the N (1 ≤ N ≤ 3,402) available charms. Each charm i in the supplied list has a weight Wi (1 ≤ Wi ≤ 400), a ‘desirability’ factor Di (1 ≤ Di ≤ 100), and can be used at most once. Bessie can only support a charm bracelet whose weight is no more than M (1 ≤ M ≤ 12,880).</p><p>Given that weight limit as a constraint and a list of the charms with their weights and desirability rating, deduce the maximum possible sum of ratings.</p><p>有 $N$ 件物品和一个容量为 $M$ 的背包。第 $i$ 件物品的重量是 $W_i$，价值是 $D_i$。求解将哪些物品装入背包可使这些物品的重量总和不超过背包容量，且价值总和最大。</p><h2 id="输入格式"><a href="#输入格式" class="headerlink" title="输入格式"></a>输入格式</h2><p>* Line 1: Two space-separated integers: N and M</p><p>* Lines 2..N+1: Line i+1 describes charm i with two space-separated integers: Wi and Di</p><p>第一行：物品个数 $N$ 和背包大小 $M$。</p><p>第二行至第 $N+1$ 行：第 $i$ 个物品的重量 $W_i$ 和价值 $D_i$。</p><h2 id="输出格式"><a href="#输出格式" class="headerlink" title="输出格式"></a>输出格式</h2><p>* Line 1: A single integer that is the greatest sum of charm desirabilities that can be achieved given the weight constraints</p><p>输出一行最大价值。</p><h2 id="样例-1"><a href="#样例-1" class="headerlink" title="样例 #1"></a>样例 #1</h2><h3 id="样例输入-1"><a href="#样例输入-1" class="headerlink" title="样例输入 #1"></a>样例输入 #1</h3><figure class="highlight plaintext"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br></pre></td><td class="code"><pre><span class="line">4 6</span><br><span class="line">1 4</span><br><span class="line">2 6</span><br><span class="line">3 12</span><br><span class="line">2 7</span><br></pre></td></tr></table></figure><h3 id="样例输出-1"><a href="#样例输出-1" class="headerlink" title="样例输出 #1"></a>样例输出 #1</h3><figure class="highlight plaintext"><table><tr><td class="gutter"><pre><span class="line">1</span><br></pre></td><td class="code"><pre><span class="line">23</span><br></pre></td></tr></table></figure><p>##代码</p><figure class="highlight plaintext"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br></pre></td><td class="code"><pre><span class="line">#include <iostream></span><br><span class="line">#include <ctime></span><br><span class="line">#include <string.h></span><br><span class="line">using namespace std;</span><br><span class="line">#define max(a, b) a > b ? a : b</span><br><span class="line">int n, m, w[3407], v[3407], f[12885];</span><br><span class="line">int main()</span><br><span class="line">{</span><br><span class="line">#ifdef LOCALFLAG</span><br><span class="line"> freopen("in.txt", "rt", stdin);</span><br><span class="line"> freopen("out.txt", "wt", stdout);</span><br><span class="line"> auto t1 = clock();</span><br><span class="line">#endif</span><br><span class="line"></span><br><span class="line"> cin >> n >> m;</span><br><span class="line"> for (int i = 1; i <= n; i++)</span><br><span class="line"> cin >> w[i] >> v[i];</span><br><span class="line"> for (int j = 0; j <= n; j++)</span><br><span class="line"> {</span><br><span class="line"> for (int i = m; i >= w[j]; i--)</span><br><span class="line"> {</span><br><span class="line"> f[i] = max(f[i], f[i - w[j]] + v[j]); //使用二维数组会爆栈，使用一维滚动数组</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> cout << f[m] << endl;</span><br><span class="line"></span><br><span class="line">#ifdef LOCALFLAG</span><br><span class="line"> auto t2 = clock();</span><br><span class="line"> cout << "Time cost= " << t2 - t1 << endl;</span><br><span class="line">#endif</span><br><span class="line">}</span><br></pre></td></tr></table></figure>]]></content>
<summary type="html"><h2 id="题目描述"><a href="#题目描述" class="headerlink" title="题目描述"></a>题目描述</h2><p>Bessie has gone to the mall’s jewelry store and spies a charm </summary>
<category term="算法题" scheme="http://zuqis.top/categories/%E7%AE%97%E6%B3%95%E9%A2%98/"/>
<category term="01背包（一维数组优化）" scheme="http://zuqis.top/tags/01%E8%83%8C%E5%8C%85%EF%BC%88%E4%B8%80%E7%BB%B4%E6%95%B0%E7%BB%84%E4%BC%98%E5%8C%96%EF%BC%89/"/>
</entry>
<entry>
<title>洛谷P3420[POI2005] SKA-Piggy Banks</title>
<link href="http://zuqis.top/2023/11/12/%E5%AD%98%E9%92%B1%E7%BD%90/"/>
<id>http://zuqis.top/2023/11/12/%E5%AD%98%E9%92%B1%E7%BD%90/</id>
<published>2023-11-11T16:06:51.092Z</published>
<updated>2023-11-11T16:10:24.888Z</updated>
<content type="html"><![CDATA[<h2 id="题目描述"><a href="#题目描述" class="headerlink" title="题目描述"></a>题目描述</h2><p>Byteazar the Dragon 拥有 $N$ 个小猪存钱罐。每一个存钱罐能够用相应的钥匙打开或者被砸开。Byteazar 已经将钥匙放入到一些存钱罐中。现在已知每个钥匙所在的存钱罐，Byteazar 想要买一辆小汽车，而且需要打开所有的存钱罐。然而，他想要破坏尽量少的存钱罐，帮助 Byteazar 去决策最少要破坏多少存钱罐。</p><h2 id="输入格式"><a href="#输入格式" class="headerlink" title="输入格式"></a>输入格式</h2><p>第一行包括一个整数 $N$ ($1\le N\le 1000000$)，表示 Byteazar the Dragon 拥有的存钱罐的数量。</p><p>存钱罐（包括它们对应的钥匙）从 $1$ 到 $N$ 编号。</p><p>接下来有 $N$ 行：第 $i+1$ 行包括一个整数 $x$，表示第 $i$ 个存钱罐对应的钥匙放置在了第 $x$ 个存钱罐中。</p><h2 id="输出格式"><a href="#输出格式" class="headerlink" title="输出格式"></a>输出格式</h2><p>仅一行：包括一个整数，表示能打开所有存钱罐的情况下，需要破坏的存钱罐的最少数量。</p><h2 id="样例-1"><a href="#样例-1" class="headerlink" title="样例 #1"></a>样例 #1</h2><h3 id="样例输入-1"><a href="#样例输入-1" class="headerlink" title="样例输入 #1"></a>样例输入 #1</h3><figure class="highlight plaintext"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br></pre></td><td class="code"><pre><span class="line">4</span><br><span class="line">2</span><br><span class="line">1</span><br><span class="line">2</span><br><span class="line">4</span><br></pre></td></tr></table></figure><h3 id="样例输出-1"><a href="#样例输出-1" class="headerlink" title="样例输出 #1"></a>样例输出 #1</h3><figure class="highlight plaintext"><table><tr><td class="gutter"><pre><span class="line">1</span><br></pre></td><td class="code"><pre><span class="line">2</span><br></pre></td></tr></table></figure><hr><p>##思路</p><p>如果这是一个联通块的“祖先”（也可以称为根吧）那么就把它砸破，这样整个联通块就都可以取出；</p><p>如果不是的话，就不管它；</p><p>从后面句话就可以很明显的推出他们之间有从属关系（如果要砸的数目最小），这是并查集的思想，是吧；</p><p>##代码</p><figure class="highlight plaintext"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br></pre></td><td class="code"><pre><span class="line">#include <iostream></span><br><span class="line">using namespace std;</span><br><span class="line">int n,a, f[1000005];</span><br><span class="line">int find(int x){</span><br><span class="line"> return x==f[x]?x:f[x]=find(f[x]);</span><br><span class="line">}</span><br><span class="line"></span><br><span class="line">int main(){</span><br><span class="line"> int ans = 0;</span><br><span class="line"> cin >> n;</span><br><span class="line"> for (int i = 1; i <= n;i++)</span><br><span class="line"> f[i] = i;</span><br><span class="line"> for (int i = 1; i <= n;i++){</span><br><span class="line"> cin >>a;</span><br><span class="line"> f[find(a)] = find(i);</span><br><span class="line"> }</span><br><span class="line"> for(int i = 1; i <= n; i++)</span><br><span class="line"> if(find(i) == i)</span><br><span class="line"> ans++;</span><br><span class="line"> cout << ans << endl;</span><br><span class="line"> return 0;</span><br><span class="line">}</span><br></pre></td></tr></table></figure>]]></content>
<summary type="html"><h2 id="题目描述"><a href="#题目描述" class="headerlink" title="题目描述"></a>题目描述</h2><p>Byteazar the Dragon 拥有 $N$ 个小猪存钱罐。每一个存钱罐能够用相应的钥匙打开或者被砸开。Byteaz</summary>
</entry>
<entry>
<title>洛谷P1892[BOI2003] 团伙</title>
<link href="http://zuqis.top/2023/11/11/gang/"/>
<id>http://zuqis.top/2023/11/11/gang/</id>
<published>2023-11-11T15:35:58.704Z</published>
<updated>2023-11-11T15:38:53.050Z</updated>
<content type="html"><![CDATA[<h2 id="题目描述"><a href="#题目描述" class="headerlink" title="题目描述"></a>题目描述</h2><p>现在有 $n$ 个人，他们之间有两种关系：朋友和敌人。我们知道：</p><ul><li>一个人的朋友的朋友是朋友</li><li>一个人的敌人的敌人是朋友</li></ul><p>现在要对这些人进行组团。两个人在一个团体内当且仅当这两个人是朋友。请求出这些人中最多可能有的团体数。</p><h2 id="输入格式"><a href="#输入格式" class="headerlink" title="输入格式"></a>输入格式</h2><p>第一行输入一个整数 $n$ 代表人数。</p><p>第二行输入一个整数 $m$ 表示接下来要列出 $m$ 个关系。</p><p>接下来 $m$ 行，每行一个字符 $opt$ 和两个整数 $p,q$，分别代表关系（朋友或敌人），有关系的两个人之中的第一个人和第二个人。其中 $opt$ 有两种可能：</p><ul><li>如果 $opt$ 为 <code>F</code>，则表明 $p$ 和 $q$ 是朋友。</li><li>如果 $opt$ 为 <code>E</code>，则表明 $p$ 和 $q$ 是敌人。</li></ul><h2 id="输出格式"><a href="#输出格式" class="headerlink" title="输出格式"></a>输出格式</h2><p>一行一个整数代表最多的团体数。</p><h2 id="样例-1"><a href="#样例-1" class="headerlink" title="样例 #1"></a>样例 #1</h2><h3 id="样例输入-1"><a href="#样例输入-1" class="headerlink" title="样例输入 #1"></a>样例输入 #1</h3><figure class="highlight plaintext"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br></pre></td><td class="code"><pre><span class="line">6</span><br><span class="line">4</span><br><span class="line">E 1 4</span><br><span class="line">F 3 5</span><br><span class="line">F 4 6</span><br><span class="line">E 1 2</span><br></pre></td></tr></table></figure><h3 id="样例输出-1"><a href="#样例输出-1" class="headerlink" title="样例输出 #1"></a>样例输出 #1</h3><figure class="highlight plaintext"><table><tr><td class="gutter"><pre><span class="line">1</span><br></pre></td><td class="code"><pre><span class="line">3</span><br></pre></td></tr></table></figure><h2 id="提示"><a href="#提示" class="headerlink" title="提示"></a>提示</h2><p>对于 $100%$ 的数据，$2 \le n \le 1000$，$1 \le m \le 5000$，$1 \le p,q \le n$。</p><hr><p>##思路</p><p>如果a和b是敌人，合并n+b和a，n+a和b</p><p>如果c和a是敌人，合并n+c和a，n+a和c</p><p>那么b和c就并在一起了</p><p>这样就符合了题目敌人的敌人是朋友的规则</p><p>##代码</p><figure class="highlight plaintext"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br></pre></td><td class="code"><pre><span class="line">#include <bits/stdc++.h></span><br><span class="line">using namespace std;</span><br><span class="line">int n, m, p, q, f[2010];</span><br><span class="line">char c;</span><br><span class="line">int find(int x){</span><br><span class="line"> return x == f[x]? x : f[x] = find(f[x]);</span><br><span class="line">}</span><br><span class="line">int main(){</span><br><span class="line"> int ans = 0;</span><br><span class="line"> cin >> n;</span><br><span class="line"> cin >> m;</span><br><span class="line"> for (int i = 1; i <= 2*n;i++)</span><br><span class="line"> f[i] = i;</span><br><span class="line"> while(m--){</span><br><span class="line"> cin >> c >> p >> q;</span><br><span class="line"> if(c == 'E'){</span><br><span class="line"> f[find(n + p)] = find(q);</span><br><span class="line"> f[find(n + q)] = find(p);</span><br><span class="line"> }</span><br><span class="line"> else</span><br><span class="line"> f[find(p)] = find(q);</span><br><span class="line"> }</span><br><span class="line"> for (int i = 1; i <= n;i++)</span><br><span class="line"> if(find(i) == i)</span><br><span class="line"> ans++;</span><br><span class="line"> cout << ans << endl;</span><br><span class="line"> return 0;</span><br><span class="line">}</span><br></pre></td></tr></table></figure>]]></content>
<summary type="html"><h2 id="题目描述"><a href="#题目描述" class="headerlink" title="题目描述"></a>题目描述</h2><p>现在有 $n$ 个人，他们之间有两种关系：朋友和敌人。我们知道：</p>
<ul>
<li>一个人的朋友的朋友是朋友</li</summary>
</entry>
<entry>
<title>洛谷P1536村村通</title>
<link href="http://zuqis.top/2023/11/11/%E6%9D%91%E6%9D%91%E9%80%9A/"/>
<id>http://zuqis.top/2023/11/11/%E6%9D%91%E6%9D%91%E9%80%9A/</id>
<published>2023-11-11T13:21:02.000Z</published>
<updated>2023-11-11T13:29:23.291Z</updated>
<content type="html"><![CDATA[<h2 id="题目描述"><a href="#题目描述" class="headerlink" title="题目描述"></a>题目描述</h2><p>某市调查城镇交通状况，得到现有城镇道路统计表。表中列出了每条道路直接连通的城镇。市政府 “村村通工程” 的目标是使全市任何两个城镇间都可以实现交通（但不一定有直接的道路相连，只要相互之间可达即可）。请你计算出最少还需要建设多少条道路？</p><h2 id="输入格式"><a href="#输入格式" class="headerlink" title="输入格式"></a>输入格式</h2><p>输入包含若干组测试数据，每组测试数据的第一行给出两个用空格隔开的正整数，分别是城镇数目 $n$ 和道路数目 $m$ ；随后的 $m$ 行对应 $m$ 条道路，每行给出一对用空格隔开的正整数，分别是该条道路直接相连的两个城镇的编号。简单起见，城镇从 $1$ 到 $n$ 编号。</p><p>注意：两个城市间可以有多条道路相通。</p><p><strong>在输入数据的最后，为一行一个整数 $0$，代表测试数据的结尾。</strong></p><h2 id="输出格式"><a href="#输出格式" class="headerlink" title="输出格式"></a>输出格式</h2><p>对于每组数据，对应一行一个整数。表示最少还需要建设的道路数目。</p><h2 id="样例-1"><a href="#样例-1" class="headerlink" title="样例 #1"></a>样例 #1</h2><h3 id="样例输入-1"><a href="#样例输入-1" class="headerlink" title="样例输入 #1"></a>样例输入 #1</h3><figure class="highlight plaintext"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br></pre></td><td class="code"><pre><span class="line">4 2</span><br><span class="line">1 3</span><br><span class="line">4 3</span><br><span class="line">3 3</span><br><span class="line">1 2</span><br><span class="line">1 3</span><br><span class="line">2 3</span><br><span class="line">5 2</span><br><span class="line">1 2</span><br><span class="line">3 5</span><br><span class="line">999 0</span><br><span class="line">0</span><br></pre></td></tr></table></figure><h3 id="样例输出-1"><a href="#样例输出-1" class="headerlink" title="样例输出 #1"></a>样例输出 #1</h3><figure class="highlight plaintext"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br></pre></td><td class="code"><pre><span class="line">1</span><br><span class="line">0</span><br><span class="line">2</span><br><span class="line">998</span><br></pre></td></tr></table></figure><h2 id="提示"><a href="#提示" class="headerlink" title="提示"></a>提示</h2><h4 id="数据规模与约定"><a href="#数据规模与约定" class="headerlink" title="数据规模与约定"></a>数据规模与约定</h4><p>对于 $100%$ 的数据，保证 $1 \le n < 1000$ 。</p><p>##思路</p><p>显然，城镇之间构成了一个又一个的集合</p><p>而对于集合的每一个元素（城镇）又没有特殊限定</p><p>那我们直接用并查集解决</p><p>每建一条公路，就是把两城镇所在的集合合并</p><p>显然，对于n个不同的集合，想要把它们连起来</p><p>至少需要连n−1条线</p><p>那我们求出一共有多少个集合</p><p>把集合数减去1，输出，搞定！</p><p>如何求集合的数量呢？</p><p>每个集合都有一个“祖宗”</p><p>又知道“祖宗”的序号不会超过1000</p><p>那好办，桶排序啊！</p><p>每遇到一个城镇，就把它的“祖宗”对应的下标变为1</p><p>我们可以通过路径压缩使同一个集合的城镇拥有同一个祖宗</p><p>那最后遍历桶，看看有多少被标为1的元素不就行了</p><p>注意有多组数据哦</p><p>##代码</p><figure class="highlight plaintext"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br></pre></td><td class="code"><pre><span class="line">#include <bits/stdc++.h></span><br><span class="line">using namespace std;</span><br><span class="line">int n, m, x, y, f[1005];</span><br><span class="line">bool barrel[1005];</span><br><span class="line">int find(int x){</span><br><span class="line"> return x == f[x]? x : f[x] = find(f[x]);</span><br><span class="line">}</span><br><span class="line">int main(){</span><br><span class="line"> while(scanf("%d%d", &n, &m)!= EOF){</span><br><span class="line"> if(n == 0) //输入0结束</span><br><span class="line"> break;</span><br><span class="line"> int sum = 0;</span><br><span class="line"> for (int i = 1; i <= n;i++)</span><br><span class="line"> f[i] = i;</span><br><span class="line"> while(m--){</span><br><span class="line"> scanf("%d%d", &x, &y);</span><br><span class="line"> if(find(x)!= find(y))</span><br><span class="line"> f[find(x)] = find(y);</span><br><span class="line"> }</span><br><span class="line"> for (int i = 1; i <= n;i++)</span><br><span class="line"> barrel[find(i)] = true; //入桶</span><br><span class="line"> for(int i = 1; i <= n; i++){</span><br><span class="line"> if(barrel[i]) //被标记的代表一个集合</span><br><span class="line"> sum++;</span><br><span class="line"> }</span><br><span class="line"> printf("%d\n", sum - 1);</span><br><span class="line"> memset(barrel, false, sizeof(barrel)); //清空桶</span><br><span class="line"> }</span><br><span class="line"> return 0;</span><br><span class="line">}</span><br></pre></td></tr></table></figure>]]></content>
<summary type="html"><h2 id="题目描述"><a href="#题目描述" class="headerlink" title="题目描述"></a>题目描述</h2><p>某市调查城镇交通状况，得到现有城镇道路统计表。表中列出了每条道路直接连通的城镇。市政府 “村村通工程” 的目标是使全市任何两</summary>
</entry>
<entry>
<title>洛谷P2078朋友</title>
<link href="http://zuqis.top/2023/11/11/friends/"/>
<id>http://zuqis.top/2023/11/11/friends/</id>
<published>2023-11-11T12:33:30.692Z</published>
<updated>2023-11-11T13:29:48.823Z</updated>
<content type="html"><![CDATA[<h2 id="题目背景"><a href="#题目背景" class="headerlink" title="题目背景"></a>题目背景</h2><p>小明在 A 公司工作，小红在 B 公司工作。</p><h2 id="题目描述"><a href="#题目描述" class="headerlink" title="题目描述"></a>题目描述</h2><p>这两个公司的员工有一个特点：一个公司的员工都是同性。</p><p>A 公司有 $N$ 名员工，其中有 $P$ 对朋友关系。B 公司有 $M$ 名员工，其中有 $Q$ 对朋友关系。朋友的朋友一定还是朋友。</p><p>每对朋友关系用两个整数 $(X_i,Y_i)$ 组成，表示朋友的编号分别为 $X_i,Y_i$。男人的编号是正数，女人的编号是负数。小明的编号是 $1$，小红的编号是 $-1$。</p><p>大家都知道，小明和小红是朋友，那么，请你写一个程序求出两公司之间，通过小明和小红认识的人最多一共能配成多少对情侣（包括他们自己）。</p><h2 id="输入格式"><a href="#输入格式" class="headerlink" title="输入格式"></a>输入格式</h2><p>输入的第一行，包含 $4$ 个空格隔开的正整数 $N,M,P,Q$。</p><p>之后 $P$ 行，每行两个正整数 $X_i,Y_i$。</p><p>之后 $Q$ 行，每行两个负整数 $X_i,Y_i$。</p><h2 id="输出格式"><a href="#输出格式" class="headerlink" title="输出格式"></a>输出格式</h2><p>输出一行一个正整数，表示通过小明和小红认识的人最多一共能配成多少对情侣（包括他们自己）。</p><h2 id="样例-1"><a href="#样例-1" class="headerlink" title="样例 #1"></a>样例 #1</h2><h3 id="样例输入-1"><a href="#样例输入-1" class="headerlink" title="样例输入 #1"></a>样例输入 #1</h3><figure class="highlight plaintext"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br></pre></td><td class="code"><pre><span class="line">4 3 4 2</span><br><span class="line">1 1</span><br><span class="line">1 2</span><br><span class="line">2 3</span><br><span class="line">1 3</span><br><span class="line">-1 -2</span><br><span class="line">-3 -3</span><br></pre></td></tr></table></figure><h3 id="样例输出-1"><a href="#样例输出-1" class="headerlink" title="样例输出 #1"></a>样例输出 #1</h3><figure class="highlight plaintext"><table><tr><td class="gutter"><pre><span class="line">1</span><br></pre></td><td class="code"><pre><span class="line">2</span><br></pre></td></tr></table></figure><h2 id="提示"><a href="#提示" class="headerlink" title="提示"></a>提示</h2><p>对于 $30 %$ 的数据，$N,M \le 100$，$P,Q \le 200$；</p><p>对于 $80 %$ 的数据，$N,M \le 4 \times 10^3$，$P,Q \le 10^4$；</p><p>对于 $100 %$ 的数据，$N,M \le 10^4$，$P,Q \le 2 \times 10^4$。</p><p>##代码</p><figure class="highlight plaintext"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br></pre></td><td class="code"><pre><span class="line">#include <bits/stdc++.h></span><br><span class="line">using namespace std;</span><br><span class="line">int n, m, p, q,x ,y, f[200005];</span><br><span class="line">int find(int x){</span><br><span class="line"> return x == f[x]? x : f[x] = find(f[x]);</span><br><span class="line">}</span><br><span class="line"></span><br><span class="line">int main(){</span><br><span class="line"> int sum1 = 0, sum2 = 0;</span><br><span class="line"> cin>>n>>m>>p>>q;</span><br><span class="line"> for (int i = 1; i <= n+m;i++)</span><br><span class="line"> f[i] = i;</span><br><span class="line"> while(p--){</span><br><span class="line"> cin>>x>>y;</span><br><span class="line"> if(find(x)!= find(y))</span><br><span class="line"> f[find(x)] = find(y);</span><br><span class="line"> }</span><br><span class="line"> while(q--){</span><br><span class="line"> cin>>x>>y;</span><br><span class="line"> x *= -1; //输入的是负数，要将其负号去除，才能进行查询</span><br><span class="line"> y *= -1;</span><br><span class="line"> if(find(x+n)!= find(y+n))</span><br><span class="line"> f[find(x+n)] = find(y+n);</span><br><span class="line"> }</span><br><span class="line"> for (int i = 1; i <= n;i++){ //求出n个人中，有多少人与1号人是朋友（包括1号人自己）</span><br><span class="line"> if(find(i) == find(1))</span><br><span class="line"> sum1++;</span><br><span class="line"> }</span><br><span class="line"> for (int i = n + 1; i <= n + m;i++){</span><br><span class="line"> if(find(i) == find(n + 1))</span><br><span class="line"> sum2++;</span><br><span class="line"> }</span><br><span class="line"> cout<<min(sum1, sum2)<<endl; //输出最小值</span><br><span class="line"> return 0;</span><br><span class="line">}</span><br></pre></td></tr></table></figure>]]></content>
<summary type="html"><h2 id="题目背景"><a href="#题目背景" class="headerlink" title="题目背景"></a>题目背景</h2><p>小明在 A 公司工作，小红在 B 公司工作。</p>
<h2 id="题目描述"><a href="#题目描述" class</summary>
</entry>
<entry>
<title>洛谷P1111修复公路</title>
<link href="http://zuqis.top/2023/11/11/repair-highway/"/>
<id>http://zuqis.top/2023/11/11/repair-highway/</id>
<published>2023-11-11T10:25:57.405Z</published>
<updated>2023-11-11T13:29:33.275Z</updated>
<content type="html"><![CDATA[<h2 id="题目背景"><a href="#题目背景" class="headerlink" title="题目背景"></a>题目背景</h2><p>A 地区在地震过后，连接所有村庄的公路都造成了损坏而无法通车。政府派人修复这些公路。</p><h2 id="题目描述"><a href="#题目描述" class="headerlink" title="题目描述"></a>题目描述</h2><p>给出 A 地区的村庄数 $N$，和公路数 $M$，公路是双向的。并告诉你每条公路的连着哪两个村庄，并告诉你什么时候能修完这条公路。问最早什么时候任意两个村庄能够通车，即最早什么时候任意两条村庄都存在至少一条修复完成的道路（可以由多条公路连成一条道路）。</p><h2 id="输入格式"><a href="#输入格式" class="headerlink" title="输入格式"></a>输入格式</h2><p>第 $1$ 行两个正整数 $N,M$。</p><p>下面 $M$ 行，每行 $3$ 个正整数 $x,y,t$，告诉你这条公路连着 $x,y$ 两个村庄，在时间t时能修复完成这条公路。</p><h2 id="输出格式"><a href="#输出格式" class="headerlink" title="输出格式"></a>输出格式</h2><p>如果全部公路修复完毕仍然存在两个村庄无法通车，则输出 $-1$，否则输出最早什么时候任意两个村庄能够通车。</p><h2 id="样例-1"><a href="#样例-1" class="headerlink" title="样例 #1"></a>样例 #1</h2><h3 id="样例输入-1"><a href="#样例输入-1" class="headerlink" title="样例输入 #1"></a>样例输入 #1</h3><figure class="highlight plaintext"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br></pre></td><td class="code"><pre><span class="line">4 4</span><br><span class="line">1 2 6</span><br><span class="line">1 3 4</span><br><span class="line">1 4 5</span><br><span class="line">4 2 3</span><br></pre></td></tr></table></figure><h3 id="样例输出-1"><a href="#样例输出-1" class="headerlink" title="样例输出 #1"></a>样例输出 #1</h3><figure class="highlight plaintext"><table><tr><td class="gutter"><pre><span class="line">1</span><br></pre></td><td class="code"><pre><span class="line">5</span><br></pre></td></tr></table></figure><h2 id="提示"><a href="#提示" class="headerlink" title="提示"></a>提示</h2><p>$1\leq x, y\leq N \le 10 ^ 3$，$1\leq M, t \le 10 ^ 5$。</p><p>##代码</p><figure class="highlight plaintext"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br></pre></td><td class="code"><pre><span class="line">#include <bits/stdc++.h></span><br><span class="line">using namespace std;</span><br><span class="line">int n, m, f[100005];</span><br><span class="line">struct edge{</span><br><span class="line"> int x,y,t; //使用结构体定义两个村庄和修复时间</span><br><span class="line">}e[100005];</span><br><span class="line">//找出x的根节点</span><br><span class="line">int find(int x){</span><br><span class="line"> return f[x] == x ? x : f[x] = find(f[x]);</span><br><span class="line">}</span><br><span class="line"></span><br><span class="line">bool cmp(edge a, edge b ){</span><br><span class="line"> return a.t < b.t;</span><br><span class="line">}</span><br><span class="line"></span><br><span class="line">int main(){</span><br><span class="line"> int ans = 0,sum = 0;</span><br><span class="line"> cin>>n>>m;</span><br><span class="line"> for(int i = 1; i <= n; i++)</span><br><span class="line"> f[i] = i;</span><br><span class="line"> for(int i = 1; i <= m; i++)</span><br><span class="line"> cin>>e[i].x>>e[i].y>>e[i].t;</span><br><span class="line"> sort(e+1,e+1+m,cmp);</span><br><span class="line"> for(int i = 1; i <= m; i++){</span><br><span class="line"> int a = find(e[i].x);</span><br><span class="line"> int b = find(e[i].y);</span><br><span class="line"> if(a!= b){</span><br><span class="line"> f[a] = b;</span><br><span class="line"> sum++;</span><br><span class="line"> ans = max(ans,e[i].t);</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> if(sum == n-1)</span><br><span class="line"> cout<<ans<<endl;</span><br><span class="line"> else</span><br><span class="line"> cout<<-1<<endl;</span><br><span class="line"> return 0;</span><br><span class="line">}</span><br></pre></td></tr></table></figure>]]></content>
<summary type="html"><h2 id="题目背景"><a href="#题目背景" class="headerlink" title="题目背景"></a>题目背景</h2><p>A 地区在地震过后，连接所有村庄的公路都造成了损坏而无法通车。政府派人修复这些公路。</p>
<h2 id="题目描述"><</summary>
</entry>
<entry>
<title>洛谷P1551亲戚</title>
<link href="http://zuqis.top/2023/11/11/%E4%BA%B2%E6%88%9A/"/>
<id>http://zuqis.top/2023/11/11/%E4%BA%B2%E6%88%9A/</id>
<published>2023-11-11T09:07:21.900Z</published>
<updated>2023-11-11T10:30:02.097Z</updated>
<content type="html"><![CDATA[<h2 id="题目背景"><a href="#题目背景" class="headerlink" title="题目背景"></a>题目背景</h2><p>若某个家族人员过于庞大，要判断两个是否是亲戚，确实还很不容易，现在给出某个亲戚关系图，求任意给出的两个人是否具有亲戚关系。</p><h2 id="题目描述"><a href="#题目描述" class="headerlink" title="题目描述"></a>题目描述</h2><p>规定：$x$ 和 $y$ 是亲戚，$y$ 和 $z$ 是亲戚，那么 $x$ 和 $z$ 也是亲戚。如果 $x$，$y$ 是亲戚，那么 $x$ 的亲戚都是 $y$ 的亲戚，$y$ 的亲戚也都是 $x$ 的亲戚。</p><h2 id="输入格式"><a href="#输入格式" class="headerlink" title="输入格式"></a>输入格式</h2><p>第一行：三个整数 $n,m,p$，（$n,m,p \le 5000$），分别表示有 $n$ 个人，$m$ 个亲戚关系，询问 $p$ 对亲戚关系。</p><p>以下 $m$ 行：每行两个数 $M_i$，$M_j$，$1 \le M_i,~M_j\le n$，表示 $M_i$ 和 $M_j$ 具有亲戚关系。</p><p>接下来 $p$ 行：每行两个数 $P_i,P_j$，询问 $P_i$ 和 $P_j$ 是否具有亲戚关系。</p><h2 id="输出格式"><a href="#输出格式" class="headerlink" title="输出格式"></a>输出格式</h2><p>$p$ 行，每行一个 <code>Yes</code> 或 <code>No</code>。表示第 $i$ 个询问的答案为“具有”或“不具有”亲戚关系。</p><h2 id="样例-1"><a href="#样例-1" class="headerlink" title="样例 #1"></a>样例 #1</h2><h3 id="样例输入-1"><a href="#样例输入-1" class="headerlink" title="样例输入 #1"></a>样例输入 #1</h3><figure class="highlight plaintext"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br></pre></td><td class="code"><pre><span class="line">6 5 3</span><br><span class="line">1 2</span><br><span class="line">1 5</span><br><span class="line">3 4</span><br><span class="line">5 2</span><br><span class="line">1 3</span><br><span class="line">1 4</span><br><span class="line">2 3</span><br><span class="line">5 6</span><br></pre></td></tr></table></figure><h3 id="样例输出-1"><a href="#样例输出-1" class="headerlink" title="样例输出 #1"></a>样例输出 #1</h3><figure class="highlight plaintext"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br></pre></td><td class="code"><pre><span class="line">Yes</span><br><span class="line">Yes</span><br><span class="line">No</span><br></pre></td></tr></table></figure><p>##代码</p><figure class="highlight plaintext"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br></pre></td><td class="code"><pre><span class="line">#include <bits/stdc++.h></span><br><span class="line">using namespace std;</span><br><span class="line">int n, m, p, x, y, f[10005];</span><br><span class="line">int find(int x){</span><br><span class="line"> return f[x] == x? x : f[x] = find(f[x]);</span><br><span class="line">}</span><br><span class="line">int main(){</span><br><span class="line"> cin>>n>>m>>p;</span><br><span class="line"> for (int i = 1; i <= n;i++)</span><br><span class="line"> f[i] = i;</span><br><span class="line"> for(int i = 1; i <= m;i++){</span><br><span class="line"> cin>>x>>y;</span><br><span class="line"> f[find(y)] = find(x);</span><br><span class="line"> }</span><br><span class="line"> for(int i = 1; i <= p;i++){</span><br><span class="line"> cin>>x>>y;</span><br><span class="line"> if(find(x) == find(y))</span><br><span class="line"> cout<<"Yes"<<endl;</span><br><span class="line"> else</span><br><span class="line"> cout<<"No"<<endl;</span><br><span class="line"> }</span><br><span class="line"> return 0;</span><br><span class="line">}</span><br></pre></td></tr></table></figure>]]></content>
<summary type="html"><h2 id="题目背景"><a href="#题目背景" class="headerlink" title="题目背景"></a>题目背景</h2><p>若某个家族人员过于庞大，要判断两个是否是亲戚，确实还很不容易，现在给出某个亲戚关系图，求任意给出的两个人是否具有亲戚关系。<</summary>
</entry>
<entry>
<title>洛谷P3367【模板】并查集</title>
<link href="http://zuqis.top/2023/09/17/%E5%B9%B6%E6%9F%A5%E9%9B%86/"/>
<id>http://zuqis.top/2023/09/17/%E5%B9%B6%E6%9F%A5%E9%9B%86/</id>
<published>2023-09-17T14:28:51.491Z</published>
<updated>2023-11-11T10:29:39.939Z</updated>
<content type="html"><![CDATA[<h2 id="题目描述"><a href="#题目描述" class="headerlink" title="题目描述"></a>题目描述</h2><p>如题，现在有一个并查集，你需要完成合并和查询操作。</p><h2 id="输入格式"><a href="#输入格式" class="headerlink" title="输入格式"></a>输入格式</h2><p>第一行包含两个整数 $N,M$ ,表示共有 $N$ 个元素和 $M$ 个操作。</p><p>接下来 $M$ 行，每行包含三个整数 $Z_i,X_i,Y_i$ 。</p><p>当 $Z_i=1$ 时，将 $X_i$ 与 $Y_i$ 所在的集合合并。</p><p>当 $Z_i=2$ 时，输出 $X_i$ 与 $Y_i$ 是否在同一集合内，是的输出<br> <code>Y</code> ；否则输出 <code>N</code> 。</p><h2 id="输出格式"><a href="#输出格式" class="headerlink" title="输出格式"></a>输出格式</h2><p>对于每一个 $Z_i=2$ 的操作，都有一行输出，每行包含一个大写字母，为 <code>Y</code> 或者 <code>N</code> 。</p><h2 id="样例-1"><a href="#样例-1" class="headerlink" title="样例 #1"></a>样例 #1</h2><h3 id="样例输入-1"><a href="#样例输入-1" class="headerlink" title="样例输入 #1"></a>样例输入 #1</h3><figure class="highlight plaintext"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br></pre></td><td class="code"><pre><span class="line">4 7</span><br><span class="line">2 1 2</span><br><span class="line">1 1 2</span><br><span class="line">2 1 2</span><br><span class="line">1 3 4</span><br><span class="line">2 1 4</span><br><span class="line">1 2 3</span><br><span class="line">2 1 4</span><br></pre></td></tr></table></figure><h3 id="样例输出-1"><a href="#样例输出-1" class="headerlink" title="样例输出 #1"></a>样例输出 #1</h3><figure class="highlight plaintext"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br></pre></td><td class="code"><pre><span class="line">N</span><br><span class="line">Y</span><br><span class="line">N</span><br><span class="line">Y</span><br></pre></td></tr></table></figure><h2 id="提示"><a href="#提示" class="headerlink" title="提示"></a>提示</h2><p>对于 $30%$ 的数据，$N \le 10$，$M \le 20$。</p><p>对于 $70%$ 的数据，$N \le 100$，$M \le 10^3$。</p><p>对于 $100%$ 的数据，$1\le N \le 10^4$，$1\le M \le 2\times 10^5$，$1 \le X_i, Y_i \le N$，$Z_i \in { 1, 2 }$。</p><p>###代码</p><figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br></pre></td><td class="code"><pre><span class="line"><span class="meta">#<span class="keyword">include</span> <span class="string"><bits/stdc++.h></span></span></span><br><span class="line"><span class="keyword">using</span> <span class="keyword">namespace</span> std;</span><br><span class="line"><span class="type">int</span> n, m, z, x, y, f[<span class="number">10005</span>];</span><br><span class="line"><span class="function"><span class="type">int</span> <span class="title">find</span><span class="params">(<span class="type">int</span> x)</span></span>{</span><br><span class="line"> <span class="keyword">if</span>(f[x] == x)</span><br><span class="line"> <span class="keyword">return</span> x;</span><br><span class="line"> <span class="keyword">else</span></span><br><span class="line"> <span class="keyword">return</span> f[x] = <span class="built_in">find</span>(f[x]);</span><br><span class="line">}</span><br><span class="line"><span class="function"><span class="type">int</span> <span class="title">main</span><span class="params">()</span></span>{</span><br><span class="line"> cin >> n >> m;</span><br><span class="line"> <span class="keyword">for</span> (<span class="type">int</span> i = <span class="number">1</span>; i <= n;i++)</span><br><span class="line"> f[i] = i;</span><br><span class="line"> <span class="keyword">while</span>(m--){</span><br><span class="line"> cin >> z >> x >> y;</span><br><span class="line"> <span class="keyword">if</span>(z == <span class="number">1</span>)</span><br><span class="line"> f[<span class="built_in">find</span>(y)] = <span class="built_in">find</span>(x);</span><br><span class="line"> <span class="keyword">else</span> </span><br><span class="line"> <span class="keyword">if</span>(<span class="built_in">find</span>(x) == <span class="built_in">find</span>(y))</span><br><span class="line"> <span class="built_in">printf</span>(<span class="string">"Y\n"</span>);</span><br><span class="line"> <span class="keyword">else</span></span><br><span class="line"> <span class="built_in">printf</span>(<span class="string">"N\n"</span>);</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">return</span> <span class="number">0</span>;</span><br><span class="line">}</span><br></pre></td></tr></table></figure>]]></content>
<summary type="html"><h2 id="题目描述"><a href="#题目描述" class="headerlink" title="题目描述"></a>题目描述</h2><p>如题，现在有一个并查集，你需要完成合并和查询操作。</p>
<h2 id="输入格式"><a href="#输入格式" cl</summary>
</entry>
<entry>
<title>Hello World</title>
<link href="http://zuqis.top/2023/09/16/hello-world/"/>
<id>http://zuqis.top/2023/09/16/hello-world/</id>
<published>2023-09-16T09:43:58.502Z</published>
<updated>2023-09-16T09:43:58.502Z</updated>
<content type="html"><![CDATA[<p>Welcome to <a href="https://hexo.io/">Hexo</a>! This is your very first post. Check <a href="https://hexo.io/docs/">documentation</a> for more info. If you get any problems when using Hexo, you can find the answer in <a href="https://hexo.io/docs/troubleshooting.html">troubleshooting</a> or you can ask me on <a href="https://github.com/hexojs/hexo/issues">GitHub</a>.</p><h2 id="Quick-Start"><a href="#Quick-Start" class="headerlink" title="Quick Start"></a>Quick Start</h2><h3 id="Create-a-new-post"><a href="#Create-a-new-post" class="headerlink" title="Create a new post"></a>Create a new post</h3><figure class="highlight bash"><table><tr><td class="gutter"><pre><span class="line">1</span><br></pre></td><td class="code"><pre><span class="line">$ hexo new <span class="string">"My New Post"</span></span><br></pre></td></tr></table></figure><p>More info: <a href="https://hexo.io/docs/writing.html">Writing</a></p><h3 id="Run-server"><a href="#Run-server" class="headerlink" title="Run server"></a>Run server</h3><figure class="highlight bash"><table><tr><td class="gutter"><pre><span class="line">1</span><br></pre></td><td class="code"><pre><span class="line">$ hexo server</span><br></pre></td></tr></table></figure><p>More info: <a href="https://hexo.io/docs/server.html">Server</a></p><h3 id="Generate-static-files"><a href="#Generate-static-files" class="headerlink" title="Generate static files"></a>Generate static files</h3><figure class="highlight bash"><table><tr><td class="gutter"><pre><span class="line">1</span><br></pre></td><td class="code"><pre><span class="line">$ hexo generate</span><br></pre></td></tr></table></figure><p>More info: <a href="https://hexo.io/docs/generating.html">Generating</a></p><h3 id="Deploy-to-remote-sites"><a href="#Deploy-to-remote-sites" class="headerlink" title="Deploy to remote sites"></a>Deploy to remote sites</h3><figure class="highlight bash"><table><tr><td class="gutter"><pre><span class="line">1</span><br></pre></td><td class="code"><pre><span class="line">$ hexo deploy</span><br></pre></td></tr></table></figure><p>More info: <a href="https://hexo.io/docs/one-command-deployment.html">Deployment</a></p>]]></content>
<summary type="html"><p>Welcome to <a href="https://hexo.io/">Hexo</a>! This is your very first post. Check <a href="https://hexo.io/docs/">documentation</a> for</summary>
</entry>
</feed>