Poj 3615 Cow Hurdles【Floyd】

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本文介绍了一个关于路径寻优的问题——CowHurdles。该问题关注如何帮助奶牛找到从一个站点到另一个站点的路径，使得跳跃过障碍物的最大高度最小。通过使用Floyd算法进行求解，提供了一种有效的方法来确定每一对站点间的最优路径。

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Cow Hurdles

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7404		Accepted: 3359

Description

Farmer John wants the cows to prepare for the county jumping competition, so Bessie and the gang are practicing jumping over hurdles. They are getting tired, though, so they want to be able to use as little energy as possible to jump over the hurdles.

Obviously, it is not very difficult for a cow to jump over several very short hurdles, but one tall hurdle can be very stressful. Thus, the cows are only concerned about the height of the tallest hurdle they have to jump over.

The cows' practice room has N (1 ≤ N ≤ 300) stations, conveniently labeled 1..N. A set of M (1 ≤ M ≤ 25,000) one-way paths connects pairs of stations; the paths are also conveniently labeled 1..M. Path i travels from station S_i to station E_i and contains exactly one hurdle of height H_i (1 ≤ H_i ≤ 1,000,000). Cows must jump hurdles in any path they traverse.

The cows have T (1 ≤ T ≤ 40,000) tasks to complete. Task i comprises two distinct numbers, A_i and B_i (1 ≤ A_i ≤ N; 1 ≤ B_i ≤ N), which connote that a cow has to travel from station A_i to station B_i (by traversing over one or more paths over some route). The cows want to take a path the minimizes the height of the tallest hurdle they jump over when traveling from A_i to B_i . Your job is to write a program that determines the path whose tallest hurdle is smallest and report that height.
　

Input

* Line 1: Three space-separated integers: N, M, and T
* Lines 2..M+1: Line i+1 contains three space-separated integers: S_i , E_i , and H_i
* Lines M+2..M+T+1: Line i+M+1 contains two space-separated integers that describe task i: A_i and B_i

Output

* Lines 1..T: Line i contains the result for task i and tells the smallest possible maximum height necessary to travel between the stations. Output -1 if it is impossible to travel between the two stations.

Sample Input

5 6 3

1 2 12

3 2 8

1 3 5

2 5 3

3 4 4

2 4 8

3 4

1 2

5 1

Sample Output

-1

Source

USACO 2007 November Silver

题意：奶牛们为了比赛要刻苦训练跳木桩。现在有n个木桩，并知道其中m对木桩的高度差。问奶牛们能从木桩u跳到木桩v，最少的跳跃高度是多少.有t个询问。

思路：

1、

if（map【j】【i】！=03f3f3f3f&&map【i】【k】！=0x3f3f3f3f）

map【j】【k】=min（map【j】【k】，max（map【j】【i】，map【i】【k】））

2、

跑一遍Floyd即可。

Ac代码：

#include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;
int map[500][500];
int main()
{
    int n,m,t;
    while(~scanf("%d%d%d",&n,&m,&t))
    {
        memset(map,0x3f3f3f3f,sizeof(map));
        for(int i=0;i<m;i++)
        {
            int x,y,w;
            scanf("%d%d%d",&x,&y,&w);
            map[x][y]=w;
        }
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                for(int k=1;k<=n;k++)
                {
                    if(map[j][i]!=0x3f3f3f3f&&map[i][k]!=0x3f3f3f3f)
                    map[j][k]=min(map[j][k],max(map[j][i],map[i][k]));
                }
            }
        }
        while(t--)
        {
            int x,y;
            scanf("%d%d",&x,&y);
            if(map[x][y]==0x3f3f3f3f)printf("-1\n");
            else printf("%d\n",map[x][y]);
        }
    }
}