HUT F - Constructing Roads

繼續最短路模板題額。。

只是多了一個條件就是有些路是已經修好的，那麼只要把修好的路之間的距離設爲0，再Dijkstra一遍就出答案~

 

 

                                                             Constructing

Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

 

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

 

#include <stdio.h>
#include <stdlib.h>
#include <algorithm>
#include <string.h>
using namespace std;
#define MAX 105
#define inf 0x7f7f7f
int n,m;
int mp[MAX][MAX];
int dis[MAX];
bool visit[MAX];

void prim( int st )
{
     for( int i=1 ; i<=n ; i++ ){
          visit[i] = false;
          dis[i] = mp[st][i];
     }     
     visit[st] = true;
     dis[st] = 0;
     int ans = 0;
     int tmp,k;
     for( int i=1 ; i<n ; i++ ){
          tmp = inf;
          for( int j=1 ; j<=n ; j++ )
               if( !visit[j] && tmp > dis[j] )
                   tmp = dis[ k = j ];
          ans += tmp;
          visit[k] = true;
          for( int j=1 ; j<=n ; j++ )
               if( !visit[j] && dis[j] > mp[k][j] )
                   dis[j] = mp[k][j];
     }
     printf("%d\n",ans);
}

int main()
{
    while( scanf("%d",&n) == 1 ){
           for( int i=1 ; i<=n ; i++ )
                for( int j=1 ; j<=n ; j++ )
                     scanf("%d",&mp[i][j]);
           scanf("%d",&m);
           int a,b;
           for( int i=1 ; i<=m ; i++ ){
                scanf("%d%d",&a,&b);     
                mp[a][b] = mp[b][a] = 0 ;  
           } 
           prim( 1 );
    }
    return 0;
}