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百度百科：瓶颈生成树
瓶颈生成树 ：无向图G的一颗瓶颈生成树是这样的一颗生成树，它最大的边权值在G的所有生成树中是最小的。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。（最小瓶颈生成树==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。[1-2]
命题：瓶颈生成树不一定是最小生成树。
下面是一个反例：
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.
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.
* 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.
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：给出n个农场和m条边,农场按1到n编号,现在有一人要从编号为1的农场出发到其他的农场去,求在这途中他最多需要携带的水的重量,注意他每到达一个农场,可以对水进行补给,且要使总共的路径长度最小。就是求最小生成树中的最长边。kruskal算法即可解决。
1 #define N 2005 2 #define M 10005 3 #include<iostream> 4 using namespace std; 5 #include<cstdio> 6 #include<algorithm> 7 struct Edge{ 8 int u,v,w; 9 bool operator <(Edge K) 10 const{return w<K.w;} 11 }edge[M]; 12 int mst=0,n,m,father[N],ans; 13 void input() 14 { 15 scanf("%d%d",&n,&m); 16 for(int i=1;i<=m;++i) 17 scanf("%d%d%d",&edge[i].u,&edge[i].v,&edge[i].w); 18 } 19 int find(int x) 20 { 21 return(father[x]==x?x:father[x]=find(father[x])); 22 } 23 void kruskal() 24 { 25 for(int i=1;i<=n;++i) 26 father[i]=i; 27 sort(edge+1,edge+m+1); 28 for(int i=1;i<=m;++i) 29 { 30 int f1=find(edge[i].u); 31 int f2=find(edge[i].v); 32 if(f1==f2) continue; 33 father[f2]=f1; 34 mst++; 35 if(mst==n-1) 36 { 37 ans=edge[i].w; 38 return; 39 } 40 } 41 } 42 int main() 43 { 44 input(); 45 kruskal(); 46 printf("%d",ans); 47 return 0; 48 }