POJ - 3615 Cow Hurdles (Floyd变形）

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 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

题目大意： 给你个N点M条边的图，T次询问S到E的所有路径中最大权值最小的那条路径的最大权值。 大体思路： 起点不固定， 4w次的询问prim算法一定超时。 可以采用一种Floyd算法的小小的变形找出任意两点的符合条件的权值。这样询问时时间复杂的为O(1), 处理时间复杂度为O（n^3）.

#include <iostream> #include <string> #include <algorithm> #include <cstdio> #include <cstring> #include <queue> #define INF 0x3f3f3f3f #define N 400 using namespace std; int dis[N][N]; int n, m, t; int main() { scanf("%d%d%d", &n, &m, &t); memset(dis, INF, sizeof(dis)); for(int i = 1; i <= m; i++) { int u, v, w; scanf("%d%d%d", &u, &v, &w); dis[u][v] = w; } 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]); } } } } for(int i = 0; i < t; i++) { int s, e; scanf("%d%d",&s, &e); if(dis[s][e] < INF) { printf("%d\n", dis[s][e]); } else { printf("-1\n"); } } return 0; }