Electrification Plan(最小生成树)

http://acm.sdut.edu.cn:8080/vjudge/contest/view.action?cid=50#problem/D

最小生成树模板，注意的是这里有k个发电站，它们不再需要连接其他的发电站，所以任何两个发电站之间的权值是0；

#include<stdio.h>
#include<string.h>
const int maxn = 110;
const int INF = 0x3f3f3f3f;
int map[maxn][maxn],power[maxn];
int n,k;

void prim()
{
    int dis[maxn],vis[maxn];
    int ans = 0;
    memset(vis,0,sizeof(vis));
    for(int i = 1; i <= n; i++)
        dis[i] = map[1][i];
    vis[1] = 1;

    for(int i = 1; i <= n-1; i++)
    {
        int min = INF,pos;

        for(int j = 1; j <= n; j++)
        {
            if(!vis[j] && dis[j] < min)
            {
                min = dis[j];
                pos = j;
            }
        }
        vis[pos] = 1;
        ans += min;
        for(int j =1; j <= n; j++)
        {
            if(!vis[j] && dis[j] > map[pos][j])
                dis[j] = map[pos][j];
        }
    }
    printf("%d
",ans);
}

int main()
{
    scanf("%d %d",&n,&k);

    /*for(int i = 1; i <= n; i++)
        for(int j = 1; j <= n; j++)
        {
            if(i != j)
                map[i][j] = INF;
            else map[i][j] = 0;
        }*/
    memset(power,0,sizeof(power));
    int x;

    for(int i = 0; i < k; i++)
    {
        scanf("%d",&x);
        power[x] = 1;
    }

    for(int i = 1; i <= n; i++)
    {
        for(int j = 1; j <= n; j++)
            scanf("%d",&map[i][j]);
    }

    for(int i = 1; i <= n; i++)
    {
        for(int j = 1; j <= n; j++)
        {
            if(power[i] && power[j])
                map[i][j] = 0;
        }
    }
    prim();
    return 0;
}