POJ 2395 Out of Hay （prim）

题目链接

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

Sample Output

分析：

最开始的时候题意没有太读明白，没有搞懂究竟是应该用dij求最短路径，还是应该用prim求最小生成树。最后终于知道应该是求最小生成树呢，但是呢这其中还要求出最小生成树中的最大的边距。

代码：

#include<stdio.h>
#include<iostream>
#include<string.h>

using namespace std;
#define inf 0x3f3f3f3f
int Tu[2009][2009];
int dis[2009];
int bj[2009];
int n,m;
void init()
{
    for(int i=1; i<=n; i++)
        for(int j=1; j<=n; j++)
        {
            if(i==j)
                Tu[i][j]=0;
            else
                Tu[i][j]=inf;
        }
    memset(dis,0,sizeof(dis));
    memset(bj,0,sizeof(bj));
}
int prim()
{
    int maxv=-inf;
    for(int i=1; i<=n; i++)
    {
        dis[i]=Tu[1][i];
        bj[i]=0;
    }
    dis[1]=0;
    bj[1]=1;
    int k;
    int Max=inf;
    for(int i=1; i<n; i++)
    {
        Max=inf;
        for(int j=1; j<=n; j++)
        {
            if(bj[j]==0&&Max>dis[j])
            {
                Max=dis[j];
                k=j;
            }
        }
        if(Max>maxv)
            maxv=Max;
        bj[k]=1;
        for(int j=1; j<=n; j++)
        {
            if(bj[j]==0&&dis[j]>Tu[k][j])
                dis[j]=Tu[k][j];
        }
    }
    return maxv;
}
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        init();
        int a,b,c;
        while(m--)
        {
            scanf("%d%d%d",&a,&b,&c);
            Tu[a][b]=Tu[b][a]=min(Tu[a][b],c);///处理重边问题，因为这wa了半天
        }
        int flag=prim();
        printf("%d
",flag);
    }
}

POJ 2395 Out of Hay （prim）

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