POJ2421 Constructing Roads

Constructing Roads

这道题很水，就是一个裸的最小生成树，最不过把已经连接的节点的值再设为0。

代码：

 #include<cstdio>
 #include<cstring>
 #define N 1010
 int dis[N][N],minn[N],n,m,a,b;
 bool u[N];
 int main(){
     scanf("%d",&n);
     for(int i=;i<=n;++i)
         for(int j=;j<=n;++j)
             scanf("%d",&dis[i][j]);
     scanf("%d",&m);
     for(int i=;i<=m;++i){
         scanf("%d%d",&a,&b);
         dis[a][b]=;
         dis[b][a]=;
     }
     memset(minn,0x7f,sizeof(minn));
     minn[]=;
     memset(u,,sizeof(u));
     for(int i=;i<=n;++i){
         int k=;
         for(int j=;j<=n;++j)
             if(u[j]&&(minn[j]<minn[k]))
                 k=j;
         u[k]=;
         for(int j=;j<=n;j++)
             if(u[j]&&(dis[k][j]<minn[j]))
                 minn[j]=dis[k][j];
     }
     int total=;
     for(int i=;i<=n;++i)
         total+=minn[i];
     printf("%d",total);
     return ;
 }