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：

 1 #include<cstdio>
 2 #include<cstring>
 3 #define N 1010
 4 int dis[N][N],minn[N],n,m,a,b;
 5 bool u[N];
 6 int main(){
 7     scanf("%d",&n);
 8     for(int i=1;i<=n;++i)
 9         for(int j=1;j<=n;++j)
10             scanf("%d",&dis[i][j]);
11     scanf("%d",&m);
12     for(int i=1;i<=m;++i){
13         scanf("%d%d",&a,&b);
14         dis[a][b]=0;
15         dis[b][a]=0;
16     }
17     memset(minn,0x7f,sizeof(minn));
18     minn[1]=0;
19     memset(u,1,sizeof(u));
20     for(int i=1;i<=n;++i){
21         int k=0;
22         for(int j=1;j<=n;++j)
23             if(u[j]&&(minn[j]<minn[k]))
24                 k=j;
25         u[k]=0;
26         for(int j=1;j<=n;j++)
27             if(u[j]&&(dis[k][j]<minn[j]))
28                 minn[j]=dis[k][j];
29     }
30     int total=0;
31     for(int i=1;i<=n;++i)
32         total+=minn[i];
33     printf("%d",total);
34     return 0;
35 }