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 = ;
 const int INF = 0x3f3f3f3f;
 int map[maxn][maxn],power[maxn];
 int n,k;

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

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

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

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

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

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