rate die clas sca sed mes nec multipl immediate

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：有n個農場，貝西在1號農場，要訪問其他n-1個農場，給出m條路，a b c表示a農場到b農場路程為c（兩個農場間可能有多條路）。貝西會挑選最短的路徑來訪問完這n-1個農場。 問在此過程中貝西會經過的最大邊是多大？ 題解：題目意思太蛋疼了，找最小生成樹上最長的邊。

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<iostream>
 4 #include<algorithm>
 5 using namespace std;
 6 
 7 const int maxn=10005;
 8 
 9 struct edge{
10     int u,v,cost;
11     bool operator<(const edge& i)const{
12         return cost<i.cost;
13     }
14 }es[maxn];
15 
16 int n,m;
17 int F[2*maxn];
18 
19 int Find(int a){
20     if(a!=F[a]) F[a]=Find(F[a]);
21     return F[a];
22 } 
23 
24 bool unite(int a,int b){
25     int x=Find(a),y=Find(b);
26     if(x==y) return false;
27     else { F[x]=y; return true; } 
28 }
29 
30 int Kruskal(){
31     sort(es,es+m);
32     int ans=0;
33     for(int i=0;i<m;i++) if(unite(es[i].u,es[i].v)) ans=max(ans,es[i].cost);
34     return ans;
35 }
36 
37 int main()
38 {   cin>>n>>m;
39     for(int i=1;i<=n;i++) F[i]=i;
40     for(int i=0;i<m;i++) scanf("%d%d%d",&es[i].u,&es[i].v,&es[i].cost);
41     int ans=Kruskal();
42     cout<<ans<<endl;
43 }

Out of Hay POJ - 2395