POJ2395 Out of Hay【Kruskal】

Input

Output

Sample Input

1 3 43

43

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】

原文：http://blog.csdn.net/lianai911/article/details/43088961