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.

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.

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.

3
0 990 692
990 0 179
692 179 0
1
1 2

179
好久没写最小生成树了，今天还忘写了一个判断，贴出来长点记性，一个模版题

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
#define maxn 105
int mapn[maxn][maxn],vis[maxn],low[maxn],n;
void prim()
{
    int sum = 0;
    memset(vis,0,sizeof(vis));
    int pos = 1;
    for(int i=1;i<=n;i++)
        low[i] = mapn[1][i];
    vis[1] = 1;
    for(int i=1;i<n;i++)
    {
        int minn = 1e9;
        for(int j=1;j<=n;j++)
        {
            if(!vis[j]&&minn>low[j])//找下一个点到这个集合的最小值
            {
                minn=low[j];//记下这个最小值
                pos=j;//记下这个点
            }
        }
        if(minn == 1e9)
            return;
        sum += minn;
        vis[pos] = 1;
        for(int j=1;j<=n;j++)
        {
            if(!vis[j]&&low[j]>mapn[pos][j])
                low[j] = mapn[pos][j];
        }
    }
    cout << sum << endl;
}
int main()
{
    while(cin >> n)
    {
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++)
                cin >> mapn[i][j];
        int m;
        cin >> m;
        while(m--)
        {
            int a,b;
            cin >> a >> b;
            mapn[b][a] = 0;
            mapn[a][b] = 0;
        }
        prim();
    }
    return 0;
}