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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

给出1到n个村庄和互相之间的距离，其中有些村庄之间已经通了道路，求这种情况下最少要建多长的道路形成最小生成树。

注意到样例里，990并不是最小的生成树里的一部分，但这已经建好了，所以再建179的就行，不然692+179才是最小生成树。所以要把已经建好的道路长度变成0，这样就成为了最小生成树的一部分，才能用算法解决

#include 
   
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            #define INF 0x3f3f3f3f#define MAXN 105#define Mod 10001using namespace std;int n,vis[MAXN],map[MAXN][MAXN],dis[MAXN];int prim(){ int i,j,pos,min,ans=0; memset(vis,0,sizeof(vis)); vis[1]=1,pos=1; for(i=1; i<=n; ++i) if(i!=pos) dis[i]=map[pos][i]; for(i=1; i
            
             map[pos][j]) dis[j]=map[pos][j]; } return ans;}int main(){ while(~scanf("%d",&n)&&n) { for(int i=1; i<=n; ++i) for(int j=1; j<=n; ++j) scanf("%d",&map[i][j]); int q,a,b; scanf("%d",&q); while(q--) { scanf("%d%d",&a,&b); map[a][b]=map[b][a]=0; } printf("%d\n",prim()); } return 0;}
            
           
          
         
        
       
      
     
    
   

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