By yusijia

Updated:

Contents

http://poj.org/problem?id=1861

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第一行输出最小生成树最大的边，第二行输出用到的电缆数量。，剩下的就是输出每条电
缆。


//注意，poj上给的示例输出是错的，被这个坑了好久，还以为是自己的代码错了
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
#define MAXN 15010

int n, m;
int father[MAXN];
int ranks[MAXN];
int num[MAXN];//用num数组记录用到了哪些边,index记录用到了几条边

struct Edge{
    int from;
    int to;
    int cost;
}edge[MAXN];

bool cmp(Edge e1, Edge e2)
{
    return e1.cost < e2.cost;
}

void init_set()
{
    for(int i = 0; i < MAXN; i++){
        father[i] = i;
        ranks[i] = 1;
    }
}

int find_set(int x)
{
    return father[x] == x ? father[x] : father[x] = find_set(father[x]);
}

void union_set(int x, int y)
{
    if(ranks[x] >= ranks[y]){
        ranks[x] += ranks[y];    //ranks数组存的是最小生成树已经连通的节点，所以最后只需检查任意一个x的father[x](根)的ranks值就知道是否是最小生成树了
        father[y] = x;
    }
    else{
        ranks[y] += ranks[x];
        father[x] = y;
    }
}

void kruskal()
{
    init_set();
    sort(edge, edge + m, cmp);
    int index = 0;
    for(int i = 0; i < m; i++){
        int root_x = find_set(edge[i].from);
        int root_y = find_set(edge[i].to);
        if(root_x != root_y){
            num[index++] = i;
            union_set(root_x, root_y);
        }
    }
    printf("%d\n%d\n" , edge[num[index - 1]].cost , index);
    for(int i = 0 ; i < index ; i++)
         printf("%d %d\n" , edge[num[i]].from , edge[num[i]].to);
}

int main()
{
    freopen("input.txt", "r", stdin);
    while(scanf("%d%d", &n, &m) != EOF){
        for(int i = 0; i < m; i++){
            scanf("%d%d%d", &edge[i].from, &edge[i].to, &edge[i].cost);
        }
        kruskal();
    }
    return 0;
}
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