Constructing Roads

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 72   Accepted Submission(s) : 20

Problem 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

30 990 692990 0 179692 179 011 2

 

Sample Output

179

//最小生成树的建立
 #include<stdio.h>
 #include<algorithm>
 using namespace std;
 int tree[10050];
 int buf[150][150];
 struct Edge{
     int x;
     int y;
     int cost;
 }edge[10050];
 void init(){
     for(int i=1;i<10050;i++){
         tree[i]=i;
         edge[i].x=0;
        edge[i].y=0;
        edge[i].cost=0;
     }
}
 int findx(int x){
     if(x!=tree[x]){
         x=findx(tree[x]);
     }
     return tree[x];
 }
 bool merge(int x,int y){
     int fx=findx(x);
     int fy=findx(y);
     if(fx!=fy){
         tree[fx]=fy;
         //表示加入最小生成树
         return true;
     }else{
         //已经使用过了
         return false;
     }
 }
 bool cmp(Edge a,Edge b){
     return a.cost<b.cost;
 }
 int main(){
     freopen("in.txt","r",stdin);
     int n;
     while(scanf("%d",&n)!=EOF){
         init();
         int bian=1;
         for(int i=1;i<=n;i++){
             for(int j=1;j<=n;j++){
                 scanf("%d",&buf[i][j]);
                 if(j>i){
                     edge[bian].x=i;
                    edge[bian].y=j;    
                    edge[bian].cost=buf[i][j];
                    bian++;
                 }
             }
         }
         bian=bian-1;
         sort(edge+1,edge+1+bian,cmp);
         int built;
         scanf("%d",&built);
         for(int i=0;i<built;i++){
             int a,b;
             scanf("%d%d",&a,&b);
             merge(a,b);
         }
         int ans=0;
         for(int i=1;i<=bian;i++){
             int xx=edge[i].x;
             int yy=edge[i].y;
            //用来判断xx与yy是否相连，因为edge已经是按照权值递增的序列，所以选择后一定是min的
             if(merge(xx,yy)==true){
                 ans+=edge[i].cost;
             }
         }
         printf("%d\n",ans);
         //printf("n=%d\n",n);
         //printf("bian=%d\n",bian);
     }
     return 0;
 }