POJ 3259 Wormholes （Bellman_ford算法）

题目链接：http://poj.org/problem?id=3259

 

Wormholes
Time Limit: 2000MS   Memory Limit: 65536K
Total Submissions: 45077   Accepted: 16625

Description

While exploring his many farms, Farmer John has discovered a number of amazing wormholes. A wormhole is very peculiar because it is a one-way path that delivers you to its destination at a time that is BEFORE you entered the wormhole! Each of FJ's farms comprises N (1 ≤ N ≤ 500) fields conveniently numbered 1..N, M (1 ≤ M ≤ 2500) paths, and W (1 ≤ W ≤ 200) wormholes.

As FJ is an avid time-traveling fan, he wants to do the following: start at some field, travel through some paths and wormholes, and return to the starting field a time before his initial departure. Perhaps he will be able to meet himself :) .

To help FJ find out whether this is possible or not, he will supply you with complete maps to F (1 ≤ F ≤ 5) of his farms. No paths will take longer than 10,000 seconds to travel and no wormhole can bring FJ back in time by more than 10,000 seconds.

Input

Line 1: A single integer, F. F farm descriptions follow. 
Line 1 of each farm: Three space-separated integers respectively: N, M, and W 
Lines 2..M+1 of each farm: Three space-separated numbers (S, E, T) that describe, respectively: a bidirectional path between S and E that requires T seconds to traverse. Two fields might be connected by more than one path. 
Lines M+2..M+W+1 of each farm: Three space-separated numbers (S, E, T) that describe, respectively: A one way path from S to E that also moves the traveler back T seconds.

Output

Lines 1..F: For each farm, output "YES" if FJ can achieve his goal, otherwise output "NO" (do not include the quotes).

Sample Input

2
3 3 1
1 2 2
1 3 4
2 3 1
3 1 3
3 2 1
1 2 3
2 3 4
3 1 8

Sample Output

NO
YES

Hint

For farm 1, FJ cannot travel back in time. 
For farm 2, FJ could travel back in time by the cycle 1->2->3->1, arriving back at his starting location 1 second before he leaves. He could start from anywhere on the cycle to accomplish this.
 
题目大意：有一个农场有n块田地，m条路连接任意两条路（无向），w个虫洞（单向）回到t秒之前，现在需要你判断她从1到n，再从n回到1是否能遇到离开前的自己
解题思路：由于题目的特殊性，只需要将穿越虫洞的时间存为负值然后用Bellman_ford算法判断是否存在负权值回路即可 如果存在说明能看到否则反之。
Bellman_ford算法  http://www.cnblogs.com/Jason-Damon/archive/2012/04/21/2460850.html
AC代码：
 1 #include <stdio.h>
 2 #include <string.h>
 3 #define inf 9999999
 4 int dis[550];
 5 int p[550][550];
 6 int n,m,w,ans;
 7 struct edge
 8 {
 9     int x,y,z;
10 }g[3000<<1];
11 void add(int s,int e,int t)
12 {
13     g[ans].x = s;
14     g[ans].y = e;
15     g[ans].z = t;
16     ans ++;
17 }
18 bool Bellman_ford()
19 {
20     int i,j;
21     for (i = 1; i <= n; i ++)
22         dis[i] = inf;
23     dis[1] = 0;
24 
25     for (i = 1; i < n; i ++)
26         for (j = 0; j < ans; j ++)
27             if (dis[g[j].y] > dis[g[j].x]+g[j].z)
28                 dis[g[j].y] = dis[g[j].x]+g[j].z;
29 
30     for (j = 0; j < ans; j ++) //遍历所有的边
31         if (dis[g[j].y] > dis[g[j].x]+g[j].z)
32             return false;
33     return true;
34 }
35 int main ()
36 {
37     int s,e,t,i,j,f;
38     scanf("%d",&f);
39     while (f --)
40     {
41         ans = 0;
42         scanf("%d%d%d",&n,&m,&w);
43         for (i = 0; i < m; i ++)
44         {
45             scanf("%d%d%d",&s,&e,&t);
46             add(s,e,t);
47             add(e,s,t);
48         }
49         for (i = 0; i < w; i ++)
50         {
51             scanf("%d%d%d",&s,&e,&t);
52             add(s,e,-t);
53         }
54         bool f = Bellman_ford();
55         if (f)
56             printf("NO\n");
57         else
58             printf("YES\n");
59     }
60     return 0;
61 }
View Code

 

posted @ 2016-09-12 20:47  gaoyanliang  阅读(207)  评论(0)  编辑  收藏  举报