Out of Hay POJ

The cows have run out of hay, a horrible event that must be remedied immediately. Bessie intends to visit the other farms to survey their hay situation. There are N (2 <= N <= 2,000) farms (numbered 1..N); Bessie starts at Farm 1. She'll traverse some or all of the M (1 <= M <= 10,000) two-way roads whose length does not exceed 1,000,000,000 that connect the farms. Some farms may be multiply connected with different length roads. All farms are connected one way or another to Farm 1.

Bessie is trying to decide how large a waterskin she will need. She knows that she needs one ounce of water for each unit of length of a road. Since she can get more water at each farm, she's only concerned about the length of the longest road. Of course, she plans her route between farms such that she minimizes the amount of water she must carry.

Help Bessie know the largest amount of water she will ever have to carry: what is the length of longest road she'll have to travel between any two farms, presuming she chooses routes that minimize that number? This means, of course, that she might backtrack over a road in order to minimize the length of the longest road she'll have to traverse.Input

* Line 1: Two space-separated integers, N and M.

* Lines 2..1+M: Line i+1 contains three space-separated integers, A_i, B_i, and L_i, describing a road from A_i to B_i of length L_i.

Output

* Line 1: A single integer that is the length of the longest road required to be traversed.

Sample Input

3 3
1 2 23
2 3 1000
1 3 43

Sample Output

43

Hint

OUTPUT DETAILS:

In order to reach farm 2, Bessie travels along a road of length 23. To reach farm 3, Bessie travels along a road of length 43. With capacity 43, she can travel along these roads provided that she refills her tank to maximum capacity before she starts down a road.

 

题意：有n个农场，贝西在1号农场，要访问其他n-1个农场，给出m条路，a  b  c表示a农场到b农场路程为c（两个农场间可能有多条路）。贝西会挑选最短的路径来访问完这n-1个农场。 问在此过程中贝西会经过的最大边是多大？

题解：题目意思太蛋疼了，找最小生成树上最长的边。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;

const int maxn=10005;

struct edge{
    int u,v,cost;
    bool operator<(const edge& i)const{
        return cost<i.cost;
    }
}es[maxn];

int n,m;
int F[2*maxn];

int Find(int a){
    if(a!=F[a]) F[a]=Find(F[a]);
    return F[a];
} 

bool unite(int a,int b){
    int x=Find(a),y=Find(b);
    if(x==y) return false;
    else { F[x]=y; return true; } 
}

int Kruskal(){
    sort(es,es+m);
    int ans=0;
    for(int i=0;i<m;i++) if(unite(es[i].u,es[i].v)) ans=max(ans,es[i].cost);
    return ans;
}

int main()
{   cin>>n>>m;
    for(int i=1;i<=n;i++) F[i]=i;
    for(int i=0;i<m;i++) scanf("%d%d%d",&es[i].u,&es[i].v,&es[i].cost);
    int ans=Kruskal();
    cout<<ans<<endl;
}