POJ 2395 Out of Hay（ 最小生成树 ）

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**链接：****传送门 **

题意：求最小生成树中的权值最大边

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/*************************************************************************
> File Name: poj2395.cpp
> Author: WArobot
> Blog: http://www.cnblogs.com/WArobot/
> Created Time: 2017年06月19日 星期一 19时00分25秒
************************************************************************/

include<stdio.h>

include<string.h>

include

using namespace std;

const int MAX_N = 2010;
const int MAX_M = 10010;
struct edge{
int from , to ,cost;
}E[MAX_M];

int n , m , par[MAX_N];

void init_union_find_set() { for(int i = 0 ; i <= n ; i++) par[i] = i; }
int find(int x) { return x == par[x] ? x : par[x] = find(par[x]); }
bool same(int x,int y) { return find(x) == find(y); }
void union_set(int x,int y) { x = find(x); y = find(y); if(x!=y) par[y] = x; }

bool cmp(edge a, edge b){
return a.cost < b.cost;
}

int Kruskal(){
init_union_find_set();
sort(E,E+m,cmp);
int ret = 0;
for(int i = 0 ; i < m ; i++){
if( !same(E[i].from , E[i].to) ){
union_set(E[i].from , E[i].to);
ret = max( ret , E[i].cost );
}
}
return ret;
}

int main(){
int from , to , cost;
while(~scanf("%d%d",&n,&m)){
for(int i = 0 ; i < m ; i++){
scanf("%d%d%d",&E[i].from,&E[i].to,&E[i].cost);
}
int ret = Kruskal();
printf("%d\n",ret);
}
return 0;
}