继续畅通工程

3
1 2 1 0
1 3 2 0
2 3 4 0
3
1 2 1 0
1 3 2 0
2 3 4 1
3
1 2 1 0
1 3 2 1
2 3 4 1
0

3
1
0

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <cmath>
using namespace std;

int per[1100];
int N, M;
struct node{
    int u, v, w;
};
node edge[110000];

int cmp(node a, node b){
    return a.w < b.w;
}

void init(){
    for(int i = 1; i <= N; ++i)
        per[i] = i;
}

int find(int x){
    if(x == per[x])
        return x;
    return per[x] = find(per[x]);
}

bool join (int x, int y){
    int fx = find(x);
    int fy = find(y);
    if(fx != fy){
        per[fx] = fy;
        return true;
    }
    return false;
}

int main (){
    while(scanf("%d", &N), N){
        M = N * (N - 1) / 2;
        int flag;
        for(int i = 0; i < M; ++i){
            scanf("%d%d%d%d", &edge[i].u, &edge[i].v, &edge[i].w, &flag);
            if(flag)
                edge[i].w = 0;
        }
        sort(edge, edge + M, cmp);
        int sum = 0;
        init();
        for(int i = 0; i < M; ++i){
            if(join(edge[i].u, edge[i].v))
                sum += edge[i].w;
        }
        printf("%d\n", sum);
    }
    return 0;
}

HDU 1879--继续畅通工程【kruskal && 最小生成树 && 水题】

原文：http://blog.csdn.net/hpuhjh/article/details/47405089