题目:poj – 2395 （B题）

The cows have run out of hay, a horrible event that mustbe remedied immediately. Bessie intends to visit the other farms to surveytheir 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 connectthe 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 thatshe needs one ounce of water for each unit of length of a road. Since she canget more water at each farm, she's only concerned about the length of thelongest road. Of course, she plans her route between farms such that sheminimizes the amount of water she must carry. 

Help Bessie know the largest amount of water she will ever have to carry: whatis 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 thelongest 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, andL_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 thelongest road required to be traversed.

Sample Input

3 3

1 2 23

2 3 1000

1 3 43

Sample Output

43

题意：从输入的数据中找最小生成树，从找到的最小生成树中的最大边；

思路可以用Kruskal 算法（首先要会并查集），也可以用prim算法

在这里我想问一下有多少人和我一样看着这道题时，题意没看懂，猜成了最短路了

下面是用 prim 算法写的

#include<stdio.h>       //我是用prim 算法写的；#include<string.h>#include<algorithm>using namespace std;#define INF 0x3f3f3f3fint e[2020][2020],book[2020],dis[2020];   //e数组存边长，book数组标记已找到的点int main()//dis 数组存从前一个点走到这个点，要走的最短边长{    int i,j,k,t,n,m;    while(~scanf("%d%d",&n,&m))  //n 为点数m为边数    {        memset(e,0,sizeof(e));        memset(book,0,sizeof(book));        memset(dis,INF,sizeof(dis));        int x1,y1,v;        for(i=0; i<m; i++)        {            scanf("%d%d%d",&x1,&y1,&v);            if(e[x1][y1]&&v>=e[x1][y1])//若之前已经输入过这条边长，又输入,要最短边长                continue;            e[x1][y1]=v;       //双向边            e[y1][x1]=v;        }        int num=1,star=1;;        dis[1]=0;        book[1]=1;        int ma=-1;//下面为pirm算法        while(num<n)     // 如找到n个点跳出循环        {            for(i=1; i<=n; i++)            {                if(!book[i]&&e[star][i]&&dis[i]>e[star][i])                    dis[i]=e[star][i];   //dis数组存从上一点走到这一点的最短边            }            int tt=INF;            for(i=1; i<=n; i++)  //找dis数组里的最短边            {                if(!book[i]&&tt>dis[i])                {                    tt=dis[i];                    star=i;                 }            }            if(ma<dis[star])    //找最小生成树里的最长边                ma=dis[star];            book[star]=1;   // 标记最短边            num++;    // 已找到点的个数        }        printf("%d\n",ma);    }    return 0;}

希望博友给提写建议，小编会努力的；