OJ 21651：：Cow Hurdles（佛罗一德的变式）

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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 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

4
8
-1

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：无非就是求个每条路的单边最大值然后取最小那个吗，由于是求任意两木桩之间的所有路径上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即为所求了。

AC代码：

#include<stdio.h>
#include<algorithm>
using namespace std;
#define INF 0x3f3f3f3f
int main()
{
int e[][];
int n,m,t,u,v,w;
scanf("%d%d%d",&n,&m,&t);
for(int i=;i<=n;i++)
for(int j=;j<=n;j++)
if(i==j)
e[i][j]=;
else
e[i][j]=INF;
for(int i=;i<=m;i++)
{
scanf("%d%d%d",&u,&v,&w);
if(e[u][v]>w)
e[u][v]=w;
}
for(int k=;k<=n;k++)
for(int i=;i<=n;i++)
for(int j=;j<=n;j++)
e[i][j]=min(e[i][j],max(e[i][k],e[k][j]));
while(t--)
{
scanf("%d%d",&u,&v);
if(e[u][v]!=INF)
printf("%d\n",e[u][v]);
else
printf("-1\n"); }
return ;
}