POJ

Constructing Roads

Time Limit: 2000MS
Memory Limit: 65536KTotal Submissions: 25252
Accepted: 11017

Description

There are N villages, which are numbered from 1 to N, and you should build some roads such that every two villages can connect to each other. We say two village A and B are connected, if and only if there is a road between A and B, or there exists a village C such that there is a road between A and C, and C and B are connected.

We know that there are already some roads between some villages and your job is the build some roads such that all the villages are connect and the length of all the roads built is minimum.

Input

The first line is an integer N (3 <= N <= 100), which is the number of villages. Then come N lines, the i-th of which contains N integers, and the j-th of these N integers is the distance (the distance should be an integer within [1, 1000]) between village i and village j.

Then there is an integer Q (0 <= Q <= N * (N + 1) / 2). Then come Q lines, each line contains two integers a and b (1 <= a < b <= N), which means the road between village a and village b has been built.

Output

You should output a line contains an integer, which is the length of all the roads to be built such that all the villages are connected, and this value is minimum.

Sample Input

30 990 692990 0 179692 179 011 2

Sample Output

179

Source

PKU Monthly,kicc


题解：
一个双向图，求最小生成树的值。注意题中给出了一部分边已连，只需要处理为权值为0即可

代码：

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<string>using namespace std;#define INF 0x3f3f3f3fint n,m;int dis[105][105],dist[105],k[105][105];int vis[105];void prime(){int sum=0;memset(vis,0,sizeof(vis));for(int i=1;i<=n;i++)dist[i]=dis[1][i];vis[1]=1;m=n-1;while(m--){int minx=INF;int p=0;for(int i=1;i<=n;i++){if(!vis[i] && minx>dist[i]){minx=dist[i];p=i;}}if(p==0)break;vis[p]=1;sum+=minx;for(int i=1;i<=n;i++){if(!vis[i] && dist[i]>dis[p][i])dist[i]=dis[p][i];}}printf("%d\n",sum);}int main(){scanf("%d",&n);memset(dist,INF,sizeof(dist));memset(dis,INF,sizeof(dis));for(int i=1;i<=n;i++){for(int j=1;j<=n;j++){scanf("%d",&dis[i][j]);}}scanf("%d",&m);for(int i=0;i<m;i++){int a,b;scanf("%d%d",&a,&b);dis[a][b]=dis[b][a]=0;}prime();return 0;}