POJ 2395 Out of Hay

Out of Hay

Time Limit: 1000MSMemory Limit: 65536KTotal Submissions: 7469Accepted: 2788

Description

The cows have run out of hay, a horrible eventthat must be remedied immediately. Bessie intends to visit theother 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 whoselength does not exceed 1,000,000,000 that connect the farms. Somefarms may be multiply connected with different length roads. Allfarms are connected one way or another to Farm1. 

Bessie is trying to decide how large a waterskin she will need. Sheknows that she needs one ounce of water for each unit of length ofa road. Since she can get more water at each farm, she's onlyconcerned about the length of the longest road. Of course, sheplans her route between farms such that she minimizes the amount ofwater she must carry. 

Help Bessie know the largest amount of water she will ever have tocarry: what is the length of longest road she'll have to travelbetween any two farms, presuming she chooses routes that minimizethat number? This means, of course, that she might backtrack over aroad in order to minimize the length of the longest road she'llhave to traverse.

Input

* Line 1: Two space-separated integers, N andM. 

* 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 lengthL_i.

Output

* Line 1: A single integer that is the length ofthe longest road required to be traversed.

Sample Input

3 31 2 232 3 10001 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. Withcapacity 43, she can travel along these roads provided that sherefills her tank to maximum capacity before she starts down aroad.

Source

USACO 2005March Silver

最小生成树，求其中的最长的路径长度

代码：

C语言: 临时自用代码

#include<stdio.h>
#include<stdlib.h>
struct point{
    int x,y,lenth;
}map[10008];
int flag[2008];
int cmp(const void *a,const void *b)
{
    struct point *c,*d;
    c=(struct point *)a;
    d=(struct point *)b;
    return c->lenth-d->lenth;
}
int father(int x)
{
    if(x==flag[x])
        return x;
    flag[x]=father(flag[x]);
    return flag[x];
}
int main()
{
    int m,n,i,max,fa,fb;
    scanf("%d%d",&n,&m);
    for(i=0;i<m;i++)
        scanf("%d%d%d",&map[i].x,&map[i].y,&map[i].lenth);
    qsort(map,m,sizeof(map[0]),cmp);
    max=0;
    for(i=0;i<=n;i++)
        flag[i]=i;
    for(i=0;i<m;i++)
    {
        fa=father(map[i].x);
        fb=father(map[i].y);
        if(fa!=fb){
            if(fa>fb)
                flag[fa]=fb;
            else flag[fb]=fa;
            if(max<map[i].lenth)
                max=map[i].lenth;
        }
    }
    printf("%d\n",max);
    return 0;
}