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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

其实就是计算连通图中是否存在负权环路，虫洞就表示负权路，正常的路是双向的，虫洞就覆盖一条单向路就行

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            #define INF 0x3f3f3f3f#define MAXN 5005#define mod 1000000007using namespace std;int n,m,w,e;int dis[MAXN];struct Node{ int beg,end,t;};Node edge[MAXN<<1];void add(int b,int en,int t){ edge[e].beg=b; edge[e].end=en; edge[e].t=t; e++;}bool relax(int p){ int t=dis[edge[p].beg]+edge[p].t; if(dis[edge[p].end]>t) { dis[edge[p].end]=t; return true; } return false;}bool bellman(){ for(int i=1;i<=n;++i) dis[i]=INF; dis[1]=0; for(int i=1; i
           
          
         
        
       
      
     
    
   

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