Wormholes  

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 83995		Accepted: 31141

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n(1 ≤ n ≤ 500)个农场，m(1 ≤ m ≤ 2500)条路, w(1 ≤ w ≤ 200)个虫洞，虫洞是单向的，而路径始终是双向的（例如有(5, 6), 则必有(6, 5)）。 问农夫是否能够通过一些路与虫洞，使得农夫能够看到以前的自己（时间倒流）。   思路：把每个顶点都入一次队，判断图中是否存在负环即可。 之前把重复的边也处理了，导致了WA多发（太惨了 删掉以后就A了 代码：

#include <iostream>
#include <queue>
#include <cstring>
#include <vector>
#include <set>
#include <algorithm>
#include <string>
#include <cstdio>
#include <cstdlib>
using namespace std;
#define MAXN 2503
#define INF 0x3f3f3f3f

int dis[MAXN];
int n, m, wei;

int h[4 * MAXN], w[4 * MAXN], e[4 * MAXN], ne[4 * MAXN], cnt[MAXN], idx;
bool vis[MAXN];

inline int read() {
    int s = 0, w = 1;
    char ch = getchar();
    while (ch<'0' || ch>'9') { if (ch == '-')w = -1; ch = getchar(); }
    while (ch >= '0'&&ch <= '9') s = s * 10 + ch - '0', ch = getchar();
    return s * w;
}
void add(int u, int v, int wei){
    e[idx] = v; w[idx] = wei; ne[idx] = h[u]; h[u] = idx ++;
    
}


bool spfa(){
    memset(cnt, 0, sizeof cnt);
    memset(vis, 0, sizeof vis);
    int cur;
    queue<int> q;
    for(int i = 1; i <= n; ++ i){
        q.push(i);
        vis[i] = true;
        cnt[i] ++;
    }
    while(!q.empty()){
        cur = q.front();
        q.pop();
        vis[cur] = false;
        for(int i = h[cur]; i != -1; i = ne[i]){
            int j = e[i];
            if(dis[j] > dis[cur] + w[i]){
                dis[j] = dis[cur] + w[i];
                
                cnt[j] ++;
                if(cnt[j] > n)
                    return true;
                if(!vis[j]){
                    q.push(j);
                    vis[j] = true;
                }
            }
        }
    }
    return false;
}

int main(){
    int t, S, E, T, W;
    for(cin >> t; t --;){
        idx = 0;
        memset(h, -1, sizeof h);
        cin >> n >> m >> W;
        for(int i = 0; i < m; ++ i){
            S = read();
            E = read();
            wei = read();
            add(S, E, wei);
            add(E, S, wei);
        }
        for(int i = 0; i < W; ++ i){
            S = read();
            E = read();
            wei = read();
            add(S, E, -wei);
        }
        
        cout << (spfa()?"YES":"NO") << endl;
                                                                                                                                                                            
    }
    return 0;
}