POJ - 2421 Constructing Roads （最小生成树 prim）

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

题意：就是有多个城市，给你每个城市之间的距离，让你修一些路，将这些城市连接起来，告诉你哪些路是已经修好的，让你算，最少修多少路，把所有的城市相连。

最小生成树，用的prim算法，随便选一个点，把与其相连的点全部进优先队列，每次读出最小的那条边，并且标记掉，读边的时候，要记录这个点是否被我读过了。

#include
#include
#include
#include
#include
#define N 99999999
using namespace std;
struct node{
    int place;
    int v;
    friend bool operator <(node n1,node n2){
        if(n1.v==n2.v)
            return n1.place > n2.place;
        else
            return n1.v>n2.v;
    }
}D,d;//优先队列
int main(){
    int Map[105][105];
    int book[105];
    int i,j,n;
    int Q,a,b;
    int sum,cas;
    while(scanf("%d",&n)!=EOF){
        for(i=1; i<=n; i++) {
            for(j=1; j<=n; j++) {
                scanf("%d",&Map[i][j]);//读入边

            }
  }
      //  printf("*\n");
        scanf("%d",&Q);
        while(Q--) {
            scanf("%d%d",&a,&b);
            Map[a][b]=0;
            Map[b][a]=0;//把修好的路赋值为0
        }
        priority_queueq;
        memset(book,0,sizeof(book));
        while(!q.empty()) {
            q.pop();
        }
        i=1;
        book[i]=1;//以这个点 已经读过了
        for(j=1; j<=n; j++) {
            if(j!=i && Map[i][j]!=N){
                D.place=j;
                D.v=Map[1][j];
                q.push(D);
                Map[i][j]=N;
                Map[j][i]=N;//读过的边都标记一下
            }
        }

        cas=1;
        sum=0;
        while(!q.empty()) {
            D=q.top();
            q.pop();
           // printf("C D.place=%d v=%d\n",D.place,D.v);
            if(book[D.place])
                continue;//如果这个点读过了 ，就不读
            sum+=D.v;
            cas++;
            book[D.place]=1;
         //   printf("cas=%d sum=%d\n",cas,sum);
            if(cas==n)
                break;
            i=D.place;
            for(j=1; j<=n; j++) {
                if(j!=i&&Map[i][j]!=N){//因为自己到自己的距离也是0，要避免这种情况
                    d.place=j;
                    d.v=Map[i][j];
                    q.push(d);
                  //  printf("R d.place=%d v=%d\n",d.place,d.v);
                    Map[i][j]=N;
                    Map[j][i]=N;
                }
            }

        }
        printf("%d\n",sum);
    }

}