Wormholes(Bellman-ford)

Wormholes

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.

Source

USACO 2006 December Gold

 1 #include<stdio.h>
 2 #include<string.h>
 3 #define MAX 0x3f3f3f3f
 4 struct path
 5 {
 6     int u , v , t ;
 7 }pa[6000];
 8
 9 int d[6000] ;
10 int n , m , w ;
11 int s , e , t ;
12 int f ;
13 int cnt ;
14
15 bool Bellman_ford ()
16 {
17     for (int i = 1 ; i <= n ; i++)
18         d[i] = MAX ;
19     d[1] = 0 ;
20     bool flag ;
21     for (int i = 1 ; i <= n ; i++) {// ‘ = ‘ 不能省
22         flag = 1 ;
23         for (int j = 0 ; j < cnt ; j++) {
24             if (d[pa[j].v] > d[pa[j].u] + pa[j].t) {
25                 flag = 0 ;
26                 d[pa[j].v] = d[pa[j].u] + pa[j].t ;
27             }
28         }
29         if (flag)
30             return true ;
31     }
32     return false ;
33 }
34
35 int main ()
36 {
37     //freopen ("a.txt" , "r" , stdin) ;
38     scanf ("%d" , &f) ;
39     while (f--) {
40         cnt = 0 ;
41         scanf ("%d%d%d" , &n , &m , &w) ;
42         for (int i = 0 ; i < m ; i++) {
43             scanf ("%d%d%d" , &s , &e , &t) ;
44             pa[cnt].u = s , pa[cnt].v = e , pa[cnt].t = t ;
45             cnt++ ;
46             pa[cnt].u = e , pa[cnt].v = s , pa[cnt].t = t ;
47             cnt++ ;
48         }
49         for (int i = 0 ; i < w ; i++ , cnt++) {
50             scanf ("%d%d%d" , &s , &e , &t) ;
51             pa[cnt].u = s , pa[cnt].v = e , pa[cnt].t = -t ;
52         }
53         if (Bellman_ford())
54             puts ("NO") ;
55         else
56             puts ("YES") ;
57     }
58     return 0 ;
59 }