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POJ 3615 Cow Hurdles （Floyd算法）

2024-11-04 14:16:02   204人阅读

Cow Hurdles
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 6142 Accepted: 2752

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 itravels from station Si to station Ei and contains exactly one hurdle of height Hi (1 ≤ Hi ≤ 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, Ai and Bi (1 ≤ Ai ≤ N; 1 ≤ Bi ≤ N), which connote that a cow has to travel from station Ai to station Bi (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 Ai to Bi . 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: Si , Ei , and Hi 
* Lines M+2..M+T+1: Line i+M+1 contains two space-separated integers that describe task i: Ai and Bi

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

4
8
-1

Source

USACO 2007 November Silver



题意：有一头牛，要进行跳木桩训练，已知有n个木桩，而且知道m对木桩之间的高度差。但是它很懒，它想尽可能的跳最小的高度就完成从任意一个木桩到任意一个木桩的跳跃，给m对点，问是否存在最小的跳跃高度使得其能够完成跳跃，如果有就输出最小高度；否则输出-1。


解析：现在要记录路径“长度”，实际上是最大跳跃高度，说白了就是，这条路径上所经过的相邻两木桩之间的差值的最大值，因为只要能跳过这个高度差最大的，高度差小的当然能跳过去了。由于是求任意两木桩之间的最大高度差值的最小值，所以我们可以用Floyd算法，对其进行处理，处理之后得到的最终结果就是最大高度的最小值了。




AC代码：

#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;
#define INF 123456789
int a[302][302];        //最大高度的最小值矩阵

int main(){
    int n, m, t;
    int x, y, w;
    while(scanf("%d%d%d", &n, &m, &t)!=EOF){
        for(int i=1; i<=n; i++)                 //初始化
            for(int j=1; j<=n; j++) a[i][j] = i==j ? 0 : INF;
        for(int i=1; i<=m; i++){                //读入高度差
            scanf("%d%d%d", &x, &y, &w);
            a[x][y] = min(a[x][y], w);          //更新最大高度差
        }
        for(int k=1; k<=n; k++)                 //Floyd
            for(int i=1; i<=n; i++)
                for(int j=1; j<=n; j++){
                    a[i][j] = min(a[i][j], max(a[i][k], a[k][j]));      
                }
        for(int i=1; i<=t; i++){
            scanf("%d%d", &x, &y);
            printf("%d\n", a[x][y]==INF ? -1 : a[x][y]);       //输出，如果还是INF，那就代表不可达，两者时之间没有路径满足要求
        }
    }
    return 0;
}






POJ 3615 Cow Hurdles （Floyd算法）


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