TOJ 1638 Out of Hay / prim

时间限制(普通/Java):1000MS/10000MS     运行内存限制:65536KByte

 

描述

 

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.

 

输入

 

* 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.

 

输出

 

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

 

样例输入

 

3 3

1 2 23

2 3 1000

1 3 43

样例输出

 

43

开始写成了最短路迪杰斯特拉

 

说真的 不知道怎么错了

 

写成了最小生成树普里姆就对了 

 

还是得思考一下

 

#include <stdio.h>  
#include <string.h>  
#define inf 0x7fffffff   
int a[2010][2010];  
int dis[2010];  
int vis[2010];  
int n,m;  
  
void prim()  
{  
    int i,j,max = 0;  
    for(i = 1; i <= n; i++)  
    {  
        dis[i] = a[1][i];  
        vis[i] = 0;  
    }  
    dis[1] = 0;  
    vis[1] = 1;  
    for(i = 1;i < n; i++)  
    {  
        int min = inf, k = 0;  
        for(j = 1; j <= n; j++)    
        {  
            if(!vis[j] && dis[j] < min)  
            {  
                min = dis[j];  
                k = j;    
            }  
        }  
        vis[k] = 1;  
        //if(max < min)  
        //  max = min;  
        for(j = 1;j <= n; j++)   
        {  
            if(!vis[j] && dis[j] > a[k][j])    
                dis[j] = a[k][j];  
        }  
    }  
    printf("%d\n",max);  
}  
int main()  
{  
    int i,j,x,y,z;  
    scanf("%d %d",&n,&m);  
    for(i = 1;i <= n; i++)  
    {  
        for(j = 1;j <= n; j++)  
        {  
            if(i == j)  
                a[i][j] = 0;  
            else  
                a[i][j] = inf;  
        }  
    }  
    for(i = 0;i < m; i++)  
    {  
        scanf("%d %d %d",&x,&y,&z);  
        if(a[x][y] > z)  
        {  
            a[x][y] = z;  
            a[y][x] = z;  
        }  
    }  
    prim();  
    return 0;  
}  

 

 

 

补充：软件开发 , C++ ,