POJ 2395 Out of Hay

Out of Hay

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 16014		Accepted: 6261

Description

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.

Source

USACO 2005 March Silver

/* 题目大意： 求一个最小生成树中的最大边 */ 
#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
int fa[2010],n,m,ans,cnt;
struct node{
    int u,v,w;
    bool operator <(const node &a)const    {
        return w<a.w;
    }
}edge[10100];
int find(int x){
    if(x==fa[x]) return fa[x];
    else return fa[x]=find(fa[x]);
}
int main()
{
    scanf("%d%d",&n,&m);
    int u,v,w;
    for(int i=1;i<=m;i++)
    {
        scanf("%d%d%d",&u,&v,&w);
        edge[i].u=u;edge[i].v=v;edge[i].w=w;
    }
    sort(edge+1,edge+m+1);
    for(int i=1;i<=n;i++)
      fa[i]=i;
    for(int i=1;i<=m;i++)
    {
        int r1=find(edge[i].u),r2=find(edge[i].v);
        if(r1==r2) continue;
        if(r1!=r2) fa[r1]=r2;
        cnt++;
        if(cnt==n-1){
            ans=edge[i].w;
            break;
        }
    }
    printf("%d\n",ans);
    return 0;
}

科普一下之瓶颈生成树：

瓶颈生成树 无向图G的一颗瓶颈生成树(bottleneck spanning tree)T是这样的一颗生成树，它最大的边权值在G的所有生成树的最大权值中是最小的。瓶颈生成树的值为T中最大权值边的权。

无向图的最小生成树一定是瓶颈生成树，但瓶颈生成树不一定是最小生成树。

命题:无向图的最小生成树一定是瓶颈生成树。

证明:可以采用反证法予以证明。

假设最小生成树不是瓶颈树，设最小生成树T的最大权边为e，则存在一棵瓶颈树Tb，其所有的边的权值小于w(e)。删除T中的e，形成两棵数T', T''，用Tb中连接T', T''的边连接这两棵树，得到新的生成树，其权值小于T，与T是最小生成树矛盾。

命题:瓶颈生成树不一定是最小生成树。

下面是一个反例:

由红色边组成的生成树是最小瓶颈树，但并非最小生成树。