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.

题意概括：

　　有一个人有n块农田和m条连接农田的道路，走每段路都需要花费一定的时间，同时他还非常热衷于时间穿梭。他想要到达某一块农田后能够通过虫洞穿越回他从起点出发之前的时间。如果他能够实现这个愿望就输出YES，否则就输出NO。

解题分析：

　　在数据中一般的道路权值为正，虫洞的权值为负。这道题就是判断是否存在负权环，若存在负权环那么每次的循环，整条路的总权值是在减少的所以最后一定会回到他出发之前的时间。

AC代码：

 1 #include<stdio.h>
 2 #include<string.h>
 3
 4 #define inf 99999999
 5 #define N 6000
 6
 7 int main()
 8 {
 9     int i, j, n, m, check, T, h, flag, k;
10     int dis[N], u[N], v[N], w[N];
11     scanf("%d", &T);
12     while(T--){
13         flag = 0;
14         scanf("%d%d%d", &n, &m, &h);
15         for(i = 1; i <= 2*m; i += 2){
16             scanf("%d%d%d", &u[i], &v[i], &w[i]);
17             u[i+1] = v[i];
18             v[i+1] = u[i];
19             w[i+1] = w[i];
20         }
21         for(i = 2*m+1; i <= 2*m+h; i++){
22             scanf("%d%d%d", &u[i], &v[i], &w[i]);
23                 w[i] = -w[i];
24         }
25
26         for(i = 1; i <= n; i++)
27             dis[i] = inf;
28         dis[1] = 0;
29         for(j = 1; j <= n-1; j++){
30             check = 1;
31             for(i = 1; i <= 2*m+h; i++){
32                 if(dis[v[i]] > dis[u[i]] + w[i]){
33                     dis[v[i]]  = dis[u[i]] + w[i];
34                     check = 0;
35                 }
36             }
37             if(check)
38                 break;
39         }
40         for(i = 1; i <= 2*m+h; i++)
41             if(dis[v[i]] > dis[u[i]] + w[i]){
42                 flag = 1;
43                 break;
44             }
45         if(flag)
46             printf("YES\n");
47         else
48             printf("NO\n");
49     }
50     return 0;
51 }