POJ

Out of Hay

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 17494 Accepted: 6885

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 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 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. With capacity 43, she can travel along these roads provided that she refills her tank to maximum capacity before she starts down a road.

题意：在保证最小生成树的前提下，找到生成树中权值最大的值，并将其输出来。

 

解题思路：可以用krusal和prim两种算法，我用的是prim 算法，我们只需要在更新sum（代表任意生成树的权值），即向树中添加新边时用一个max函数求出最大值即可。

代码如下：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define INF 0x3f3f3f3f
#define N 3000
int e[N][N],b[N],n,m,dis[N];
int prim()
{
    int max1=-1,i,j,k,s=0,cou=0,minn;
    memset(b,0,sizeof(b));
    for(i=1; i<=n; i++)
        dis[i]=e[1][i];
    b[1]=1;
    cou++;
    while(cou<n)
    {
        minn=INF;//这个注意位置，别放错了
        for(i=1; i<=n; i++)
        {

           //min=INF;放这找的是最后一个小的点,错错错！！！！！！！
            if(!b[i]&&dis[i]<minn)
            {
                minn=dis[i];
                j=i;
            }
        }
        b[j]=1;
        cou++;
        //s+=dis[j];//加的是进入树的变得权值
        max1=max(minn,max1);//每条边得最大权值
        for(k=1; k<=n; k++)
        {
            if(!b[k]&&dis[k]>e[j][k])
                dis[k]=e[j][k];
        }
    }
    return max1;
}
int main()
{
    while(~scanf("%d %d",&n,&m))
    {
        int u,v,w;
        for(int i=1; i<=n; i++)
        {
            for(int j=1; j<=n; j++)
            {
                if(i==j)
                    e[i][j]=0;
                else
                    e[i][j]= INF;
            }
        }
        for(int i=1; i<=m; i++)
        {
            scanf("%d %d %d",&u,&v,&w);
            if(e[u][v]>w)//注意这里是易错的地方，题上有提示，即到同一点时有多条路，我们肯定选最小的了
                e[u][v]=e[v][u]=w;
        }
        printf("%d\n",prim());
    }
    return 0;
}