POJ 2421 有一条连通下的最小生成树

Constructing Roads

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 22440		Accepted: 9588

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 <iostream>
#include <cstdio>
#include <algorithm>
#include <string.h>
#include <climits>
using namespace std;


const int N = 110;
int father[N];
struct edge{
    int lp,rp,value;
}ee[N*N];
int map[N][N],flag[N][N],numedge,n;
bool cmp(edge a,edge b){
    return a.value < b.value;
}

//并查集模板
int find(int x){
    if(father[x] == x)
        return father[x];
       else
            father[x]=find(father[x]);
    return father[x];
}
int merge(int x,int y){
    int fx = find(x);
    int fy = find(y);
    if(fx!=fy)
    {
      father[fy] = fx;
      return 1;
    }
    else
        return 0;  //要么bool类型，返回-1也是ture,只有return 0 才是false

}

//克鲁斯卡尔算法
int kruskal(){
    sort(ee,ee+numedge,cmp);
    for(int i = 1; i <= n; ++i)
        father[i] = i;
    int sum = 0;
    for(int i = 0; i < numedge; ++i){
        int lx = ee[i].lp;
        int rx = ee[i].rp;
        if(merge(lx,rx)){
            sum += ee[i].value;
        }
    }
    return sum;
}
int main(){
    //freopen("1.txt","r",stdin);
    while(scanf("%d",&n) != EOF){
      memset(flag,0,sizeof(flag));
        for(int i = 1; i <= n; ++i){
          for(int j = 1; j <= n; ++j)
              scanf("%d",&map[i][j]);
        }
        int m,x,y;
        scanf("%d",&m);
        while(m--){
          scanf("%d%d",&x,&y);
          flag[x][y] = 1;
        }
        numedge = 0;
        //无向图，只用放1-2,1-3,2-3
        for(int i = 1; i < n; ++i){
            for(int j = i + 1; j <= n; j++){
                if(flag[i][j]){
                    ee[numedge].lp = i;
                    ee[numedge].rp = j;
                    ee[numedge].value = 0;
                    numedge++;
                }
                else{
                    ee[numedge].lp = i;
                    ee[numedge].rp = j;
                    ee[numedge].value = map[i][j];
                    numedge++;
                }
            }
        }
        int ans = kruskal();
        printf("%d
",ans);
    }
    return 0;
}

坑爹，开始在并查集中找到共同根后应该返回false，自己写成return -1，返回的还是true，找错找了半天