POJ

问题描述
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.

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 31 2 232 3 10001 3 43

Sample Output

43

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分析：

题目刚开始没读懂题啊，还以为搞个最短路，后来wa了，才知道题目不是这么玩的，原来是走完所有的farm，所以是最小生成树了，并且要求的是最小生成树里的最长一条路径，则用kruskal实时更新最大的一条路径就行。

代码如下：

#include<cstdio>#include<algorithm>using namespace std;const int maxn = 1000+10;const int maxe = 20000+10;struct edge{    int u, v, cost;}es[maxe];int cmp(const edge &a, const edge &b){    return a.cost < b.cost;}int N,M;int par[maxn];void Init(){    for(int i=1; i<=N; i++)    {        par[i] = i;    }}int find(int x){    int r = x;    while(r!=par[r])    r = par[r];    int i = x;    int j;      while(i != r)    {        j = par[i];        par[i] = r;        i = j;    }    return r;}bool same(int x, int y){    return find(x)==find(y);}void unite(int x, int y){    par[find(x)] = find(y);}int kruskal(){    sort(es, es+M, cmp);    Init();    int res = 0;    for(int i=0; i<M; i++)    {        edge e = es[i];        if(!same(e.u, e.v))        {            res = max(res, e.cost);            unite(e.u, e.v);        }    }    return res;}int main(){    scanf("%d%d",&N,&M);    for(int i=0; i<M; i++)    {        scanf("%d%d%d",&es[i].u,&es[i].v,&es[i].cost);    }    kruskal();    printf("%d\n", kruskal());    return 0;}