题目链接： http://acm.hdu.edu.cn/showproblem.php?pid=1102

Constructing Roads

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K
(Java/Others)
Total Submission(s): 14983 Accepted Submission(s): 5715

Problem 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

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

Sample Output

179

#include <iostream>
#include <stdio.h>
#include <math.h>
#include <algorithm>
#include <string.h>

using namespace std;

int n,m;
int p[1005][1005];
int low [1005];
int vis[1005];

int prim ()
{
    int vis[1000];
    memset(vis,0,sizeof(vis));

    int low [1000],res=0,minn;
    int pos=1;
    vis[1]=1;

    for (int i=1;i<=n;i++)
    {
        if (i!=pos)
            low [i]=p[pos][i];
    }

    for (int j=1;j<n;j++)
    {
        minn=1000000;
        for (int i=1;i<=n;i++)
        {
            if ( !vis[i] && minn>low[i] )
            {
                pos=i;
                minn=low[i];
            }
        }

        res+=minn;
        vis[pos]=1;

        for (int i=1;i<=n;i++)
        {
            if ( !vis[i] && low[i]>p[pos][i]  )
            {
                low[i]=p[pos][i];
            }
        }
    }
    return res;
}
int main ()
{
    int kkk;
    while (scanf("%d",&n)!=EOF)
    {
        memset(p,100000,sizeof(p));

        for (int i=1;i<=n;i++)
        {
                for (int j=1;j<=n;j++)
            {
                scanf ("%d",&kkk);
                    p[i][j]=kkk;
            }
        }
        int c,d;
        scanf ("%d",&m);
        while (m--)
        {
            scanf ("%d%d",&c,&d);
            p[c][d]=p[d][c]=0;
        }
        int tt=prim();
        printf("%d\n",tt);
    }
    return 0;
}