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Prim.cpp
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61 lines (49 loc) · 1.24 KB
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#include<bits/stdc++.h>
using namespace std;
#define N 505
#define inf 99999999
int e[N][N], dis[N], v[N];
int main()
{
int n, m, Min, t1, t2, t3, j, pos;
//输入点数、边数
while(scanf("%d%d", &n, &m)) {
memset(v,0,sizeof(v));
memset(dis,0,sizeof(dis));
int cnt = 0, sum = 0; //点数、路径权值和
//m = n * (n-1) / 2;
//初始化图
for(int i=1; i<=n; i++){
for(int j=1; j<=n; j++){
if(i == j) e[i][j] = 0;
else e[i][j] = inf;
}
}
//无向图建边
for(int i=1; i<=m; i++){
scanf("%d %d %d",&t1, &t2, &t3);
e[t1][t2] = t3;
e[t2][t1] = t3;
}
//初始化dis数组,1号顶点到各个顶点初始距离
for(int i=1; i<=n; i++)
dis[i] = e[1][i];
v[1] = 1; //第一个结点访问了
for(int i=1; i<n; i++){ //访问了第一个以后,剩下n-1个结点
Min = inf;
for(int j=1; j<=n; j++){
if(v[j]==0 && dis[j] < Min){ //找出最小距离的节点
Min = dis[pos = j];
}
}
v[pos] = 1;
sum += Min;
for(int j=1 ;j<=n; j++){
if(v[j]==0)
dis[j] = min(dis[j],e[pos][j]);
}
}
printf("%d\n",sum); //选出n-1条边,使得边的总长度之和最短
}
return 0;
}