Out of Hay（Kruska算法求最小生成树l）

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：1号农场的草被牛吃完了，Bessie必须从其他农场运草回来，总共有N个农场，Bessie要去其他所有的农场运草回来，他想要使总路程最短并且路线能连接所有的农场。须要考虑到路上带的水袋大小。因为水袋大小和路线中距离最长的两个农场之间的路有关，现在Bessie想要求出满足要求的路线中两个农场之间最长的路距离是多少。

思路：

就是求最小生成树，只不过不是求总的路长了，而是让我们求最小生成树中最长的那一条边。稍微改了改前面做的题就AC了。。。

代码：

#include <iostream>
#include <cstring>
#include <stdio.h>
#include <algorithm>
using namespace std;
struct edgenode{
int from;
int to;
int w;
}edge[40040];
int father[2100];
int find(int x)   //并查集
{
    if(x!=father[x])
        father[x]=find(father[x]);
    return father[x];
}
int cmp(edgenode a,edgenode b)  //排序按由由小到大快排
{
    return a.w<b.w;
}
void kruskal(int n,int m)  //Kruskal算法
{
    sort(edge,edge+m,cmp);
    int tot=0,k=0;//k为计数器
    for(int i=0;i<m;i++)
    {
        int u=find(edge[i].from);
        int v=find(edge[i].to);
        if(u!=v)  //不属于同一并查集
        {
            if(tot<edge[i].w)
                tot=edge[i].w;
            father[v]=u;  //让其进入最大生成树
            k++;
            if(k==n-1)break;  //如果k=n-1说明最小生成树已经生成则break;
        }
    }
    cout<<tot<<endl;
}
int main()
{
    int n,m;
    int i;
    while(cin>>n>>m)
    {
        for(i=1;i<=n;i++) father[i]=i;  //初始化并查集
        for(i=0;i<m;i++)
        {
            cin>>edge[i].from>>edge[i].to>>edge[i].w;
        }
        kruskal(n,m);
    }


}