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

题意：
     有N个村庄，编号从1到N。现需要在这N个村庄之间修路，使得任何两个村庄之间都可以连通。称A、B两个村庄是连通的，
当且仅当A与B有路直接连接，或者存在村庄C，使得A和C两村庄之间有路连接，且C和B之间有路连接。已知某些村庄之间已经有
路直接连接了，试修建一些路使得所有村庄都是连通的、且修路总长度最短。

#include <cstdio>

#include <iostream>

#include<cstring>

#include<algorithm>

#include<cmath>

using namespace std;

const int MAXN=;

 int p[MAXN];

 bool sum[MAXN];

 int m[MAXN][MAXN];

struct node

 {

     int x,y,l;

 }a[];

 bool cmp(node a,node b)

 {

     return a.l<b.l;

 }

 int Find(int x)

 {

     return x==p[x]?x:(p[x]=Find(p[x]));

 }

int  Union(int  R1,int R2)

 {

     int r1=Find(R1);

     int r2=Find(R2);

     if(r1!=r2)

     {

         p[r1]=r2;

        return ;

     }

     else return ;

 }

int main()

{

    int n;

    int cnt=,i,j;

 while(~scanf("%d",&n))

   {

       cnt =;

       memset(sum,,sizeof(sum));

       for(i=;i<=n;i++)

           p[i]=i;

        for(i=;i<=n;i++)

            for( j=;j<=n;j++)

              scanf("%d",&m[i][j]);

          int t,c,b;

        scanf("%d",&t);

        while(t--)                  //将已经修好的路长度清零

        {

            scanf("%d%d",&c,&b);

            m[c][b]=m[b][c]=;

        }

        int k=;

     for(i=;i<=n;i++)

     {

         for(j=+i;j<=n;j++)

            {

                a[k].x=i;

                a[k].y=j;

                a[k].l=m[i][j];

                    k++;

            }

     }

 sort(a,a+k,cmp);

      for(i=;i<k;i++)

     {

         if(Union(a[i].x,a[i].y)==)

            cnt+=a[i].l;

}

     printf("%d\n",cnt);

   }

    return ;

}