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<a href="/2023/11/16/%E6%B8%B8%E8%89%87%E9%97%AE%E9%A2%98/" class="post-title-link" itemprop="url">洛谷P1359租用游艇</a>
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<time title="创建时间：2023-11-16 22:48:15 / 修改时间：23:15:46" itemprop="dateCreated datePublished" datetime="2023-11-16T22:48:15+08:00">2023-11-16</time>
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<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>
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<a href="/2023/11/14/p1466%E9%9B%86%E5%90%88/" class="post-title-link" itemprop="url">洛谷P1466[USACO2.2] 集合 Subset Sums</a>
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<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>
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<a href="/2023/11/14/01%E8%83%8C%E5%8C%85/" class="post-title-link" itemprop="url">洛谷P2871[USACO07DEC] Charm Bracelet S</a>
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<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>
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<a href="/2023/11/12/%E5%AD%98%E9%92%B1%E7%BD%90/" class="post-title-link" itemprop="url">洛谷P3420[POI2005] SKA-Piggy Banks</a>
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<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>
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<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>
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<a href="/2023/11/11/%E6%9D%91%E6%9D%91%E9%80%9A/" class="post-title-link" itemprop="url">洛谷P1536村村通</a>
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<time title="创建时间：2023-11-11 21:21:02 / 修改时间：21:29:23" itemprop="dateCreated datePublished" datetime="2023-11-11T21:21:02+08:00">2023-11-11</time>
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<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>
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