POJ 3615 A - Cow Hurdles 弗洛伊德算法
来源:互联网 发布:淘宝买的单机游戏 编辑:程序博客网 时间:2024/06/05 19:14
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 ofM (1 ≤ M ≤ 25,000) one-way paths connects pairs of stations; the paths are also conveniently labeled 1..M. Pathi travels from station Si to station Ei and contains exactly one hurdle of heightHi (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 stationAi 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 fromAi 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.
5 6 31 2 123 2 81 3 52 5 33 4 42 4 83 41 25 1
深入理解floyd算法#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <stack>#include <queue>#include <set>#include <map>#include <algorithm>using namespace std;typedef pair<int, int> P;#define LOCAL#define INF 0x3f3f3f3f#define MAX_N 1005int edge[MAX_N][MAX_N];int N, M, T;void floyd(){ for(int k = 1; k <= N; k++) { for(int i = 1; i <= N; i++) { for(int j = 1; j <= N; j++) { edge[i][j] = min(edge[i][j], max(edge[i][k], edge[k][j])); } } }}int main(){#ifdef LOCALfreopen("b:\\data.in.txt", "r", stdin);#endif cin >> N >> M >> T;//N个点M条边T个请求 memset(edge, INF, sizeof(edge)); int from, to, cost; for(int i=1; i <= M; i++)//输入图 { scanf("%d%d%d", &from, &to, &cost); edge[from][to] = cost; } floyd(); for(int i = 1; i <= T; i++) { scanf("%d%d", &from, &to); if(edge[from][to] < INF) cout << edge[from][to] << endl; else cout << -1 << endl; } return 0;}
- POJ 3615 A - Cow Hurdles 弗洛伊德算法
- poj 3615(Cow Hurdles) floy 算法
- POJ 3615 Cow Hurdles(Floyd算法)
- POJ 3615 Cow Hurdles (Floyd算法)
- POJ 3615 Cow Hurdles //Floyd算法
- POJ 3615 Cow Hurdles
- poj 3615 Cow Hurdles
- poj 3615 Cow Hurdles
- poj 3615 Cow Hurdles
- poj 3615 Cow Hurdles
- poj 3615 Cow Hurdles
- POJ 3615 Cow Hurdles.
- POJ -- 3615 Cow Hurdles
- poj 3615 Cow Hurdles
- poj 3615 Cow Hurdles
- POJ--3615|Cow Hurdles
- POJ 3615 Cow Hurdles
- poj 3615 Cow Hurdles floyed
- 二叉树的层次遍历
- c++内存分配(new,operator new)详解
- How To Setup Partitioned Linux Block Devices Using UDEV (Non-ASMLIB) And Assign Them To ASM
- 回调函数(callback)是什么
- CSS学习笔记
- POJ 3615 A - Cow Hurdles 弗洛伊德算法
- java网络编程-下载二进制文件的正确流写法
- 嵌入式
- Iptables整理
- BroadcastReceiver广播接收器
- 第3周 C语言及程序设计初步例程-42 将数据输出到文本文件
- PicketLink入门指南
- PHP学习(十六)--封装性
- 十四、符号表Map(Set)的应用