poj 3615(Floyd变形)

Cow Hurdles

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 7278 Accepted: 3314

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 31 2 123 2 81 3 52 5 33 4 42 4 83 41 25 1

Sample Output

48
-1
题意：多个询问，求A->B的所有路径中最小的最大边。
解题思路：Floyd算法，主要还是理解它的思想，dis[i][j]表示从i->j最小的最大边，则根据Floyd的松弛条件，略加修改即可：dis[i][j] = min{max{dis[i][k],dis[k][j]}}
#include<iostream>#include<cstdio>#include<cstring>using namespace std;const int maxn = 305;const int inf = 0x3f3f3f3f;int n,m,q,map[maxn][maxn];int dis[maxn][maxn];void floyd(){for(int k = 1; k <= n; k++)for(int i = 1; i <= n; i++)for(int j = 1; j <= n; j++){if(dis[i][j] > max(dis[i][k],dis[k][j]))dis[i][j] = max(dis[i][k],dis[k][j]);if(dis[i][j] > map[i][j])dis[i][j] = map[i][j];}}int main(){int u,v,w;while(scanf("%d%d%d",&n,&m,&q)!=EOF){memset(dis,inf,sizeof(dis));memset(map,inf,sizeof(map));for(int i = 1; i <= n; i++) {dis[i][i] = 0;map[i][i] = 0;}for(int i = 1; i <= m; i++){scanf("%d%d%d",&u,&v,&w);map[u][v] = min(map[u][v],w);}floyd();while(q--){scanf("%d%d",&u,&v);if(dis[u][v] == inf)printf("-1\n");else printf("%d\n",dis[u][v]);}}return 0;}