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POJ2395 Out of Hay【Kruskal】

2024-11-15 20:23:39   202人阅读

Out of Hay
Time Limit: 1000MSMemory Limit: 65536K
Total Submissions: 11656Accepted: 4562
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(专横的) 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：1号农场的草被牛吃完了，Bessie必须从其他农场运草回来，总共有N个农场，Bessie要

去其他所有的农场运草回来，他想要使总路程最短并且路线能连接所有的农场。必须要考虑到路上

带的水袋大小。因为水袋大小和路线中距离最长的两个农场之间的路有关，现在Bessie想要求出满

足要求的路线中两个农场之间最长的路距离是多少。

思路：满足要求的路线其实就是最小生成树，路线中两个农场之间最长的路距离就是最小生成树上

最长的边。这样用Kruskal求最小生成树的时候，用Max求出最小生成树上最长的边。


<span style="font-family:Microsoft YaHei;font-size:18px;">#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
const int MAXN = 2100;
const int MAXM = 40040;
struct EdgeNode
{
    int from;
    int to;
    int w;
}Edges[MAXM];

int father[MAXN];

int find(int x)
{
    if(x != father[x])
        father[x] = find(father[x]);
    return father[x];
}

int cmp(EdgeNode a,EdgeNode b)
{
    return a.w < b.w;
}

void Kruskal(int N,int M)
{
    sort(Edges,Edges+M,cmp);
    int Count = 0, Max = 0;
    for(int i = 0; i < M; ++i)
    {
        int u = find(Edges[i].from);
        int v = find(Edges[i].to);
        if(u != v)
        {
            father[v] = u;
            Count++;
            if(Max < Edges[i].w)
                Max = Edges[i].w;
            if(Count == N-1)
                break;
        }
    }
    cout << Max << endl;
}
int main()
{
    int N,M;
    while(~scanf("%d%d",&N,&M))
    {
        for(int i = 1; i <= N; ++i)
            father[i] = i;
        for(int i = 0; i < M; ++i)
        {
            scanf("%d%d%d",&Edges[i].from, &Edges[i].to, &Edges[i].w);
        }
        Kruskal(N,M);
    }

    return 0;
}
</span>


POJ2395 Out of Hay【Kruskal】


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