瓶頸生成樹與最小生成樹 POJ 2395 Out of Hay

百度百科：瓶頸生成樹

瓶頸生成樹 ：無向圖G的一顆瓶頸生成樹是這樣的一顆生成樹，它最大的邊權值在G的所有生成樹中是最小的。瓶頸生成樹的值為T中最大權值邊的權。

無向圖的最小生成樹一定是瓶頸生成樹，但瓶頸生成樹不一定是最小生成樹。（最小瓶頸生成樹==最小生成樹）

命題：無向圖的最小生成樹一定是瓶頸生成樹。

證明：可以采用反證法予以證明。

假設最小生成樹不是瓶頸樹，設最小生成樹T的最大權邊為e，則存在一棵瓶頸樹Tb，其所有的邊的權值小於w(e)。刪除T中的e，形成兩棵數T', T''，用Tb中連接T', T''的邊連接這兩棵樹，得到新的生成樹，其權值小於T，與T是最小生成樹矛盾。[1-2]  

命題：瓶頸生成樹不一定是最小生成樹。

下面是一個反例：

 

由紅色邊組成的生成樹是瓶頸樹，但並非最小生成樹。

  POJ 2395 Out of Hay

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 15380		Accepted: 6008

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.

 

題意：給出n個農場和m條邊,農場按1到n編號,現在有一人要從編號為1的農場出發到其他的農場去,求在這途中他最多需要攜帶的水的重量,注意他每到達一個農場,可以對水進行補給,且要使總共的路徑長度最小。就是求最小生成樹中的最長邊。kruskal算法即可解決。

#define N 2005
#define M 10005
#include<iostream>
using namespace std;
#include<cstdio>
#include<algorithm>
struct Edge{
    int u,v,w;
    bool operator <(Edge K)
    const{return w<K.w;}
}edge[M];
int mst=0,n,m,father[N],ans;
void input()
{
    scanf("%d%d",&n,&m);
    for(int i=1;i<=m;++i)
      scanf("%d%d%d",&edge[i].u,&edge[i].v,&edge[i].w);
}
int find(int x)
{
    return(father[x]==x?x:father[x]=find(father[x]));
}
void kruskal()
{
    for(int i=1;i<=n;++i)
      father[i]=i;
    sort(edge+1,edge+m+1);
    for(int i=1;i<=m;++i)
    {
        int f1=find(edge[i].u);
        int f2=find(edge[i].v);
        if(f1==f2) continue;
        father[f2]=f1;
        mst++;
        if(mst==n-1)
        {
            ans=edge[i].w;
            return;
        }
    }
}
int main()
{
    input();
    kruskal();
    printf("%d",ans);
    return 0;
}