POJ 3615 Cow Hurdles （flyod）

Cow Hurdles

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8506		Accepted: 3751

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

4
8
-1

Source

USACO 2007 November Silver

 

题意就是，一些牛要翻栏杆，他们要从A一直到B，中间路过的点需要翻过一个高度为h的栏杆，他们为了省最多的力气，想翻栏杆的高度最低（题目认为比最高的栏杆低的栏杆，牛翻过去不费力），同时找到这个最高栏杆的高度并输出。

做法呢，其实就是把最高的栏杆当做边权，求一个最短路，看到只有300个点，直接floyd，每个点都优化一遍，然后输出就可以了。

//#include<bits/stdc++.h>
#include<cstdio>
#include<queue>
#include<algorithm>
#include<cstring>
#include<string>
#include<vector>
#include<cmath>
#include<iostream>
using namespace std;
#define maxn 305
#define INF 99999999
typedef pair<int,int> pii;
vector<pii> e[maxn];
int d[maxn],vis[maxn],mp[maxn][maxn];
int n,b,k;
int main()
{
    int n, m, t;
    while(~scanf("%d %d %d",&n,&m,&t))
    {
        for(int i=1; i<=n; i++)
            for(int j=1; j<=n; j++)
            {
                if(i==j) mp[i][j]=0;
                else mp[i][j]=INF;
            }
        for(int i=1; i<=m; i++)
        {
            int x,y,z;
            scanf("%d %d %d",&x,&y,&z);
            mp[x][y] = min(mp[x][y], z);
        }
        for(int k=1; k<=n; k++)
            for(int i=1; i<=n; i++)
                for(int j=1; j<=n; j++)
                {
                    mp[i][j] = min(mp[i][j], max(mp[i][k], mp[k][j]));
                }
        for(int i=1; i<=t; i++)
        {
            int x,y;
            scanf("%d %d",&x,&y);
           if(mp[x][y]==INF)
            printf("-1\n");
           else
            printf("%d\n",mp[x][y]);
        }
    }
    return 0;
}