# POJ 3615 Cow Hurdles （Floyd算法）

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: NM, 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 ⑴ 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

Source

USACO 2007 November Silver

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
", a[x][y]==INF ? ⑴ : a[x][y]); //输出，如果还是INF，那就代表不可达，二者时之间没有路径满足要求
}
}
return 0;
}

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