最短路

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 ：

　　有一个新概念虫洞，这个人可以穿过这条道路，并且时间会回溯T妙，问此人是否可以通过穿越虫洞回溯时间看到自己。实质上就是判断图中有无负环。

思路 ：

SPFA ： BF算法的优化，从一个点出发，更新它可以到达的所有点，把可以到的，更新的点，并且不再队列中的点，再重新加入到队列中。

　　　　如何判断有无负环呢？一个点的更新次数如果大于 n-1次，那么一定存在负环，因为一个点被加入队列两次，意味着这个点在BF 中被更新两次，那么这个点如果加入队列的次数大于 n-1 次，那么则存在负环。

代码示例 ：

const int inf = 1 << 29;
int n,m, w;
struct ed
{
    int to, cost;
    ed(int _t = 0, int _c = 0):to(_t),cost(_c){}
};
vector<ed>edge[505];
bool vis[505];
int d[505];
int cnt[505];

bool spfa(){
    queue<int>que;
    memset(cnt, 0, sizeof(cnt));
    memset(vis, false, sizeof(vis));
    for(int i = 1; i <= 500; i++) d[i] = inf;
    d[1] = 0;
    cnt[1] = 1;    
    que.push(1);
    
    while(!que.empty()){
        int u = que.front();
        que.pop();
        vis[u] = false;
        for(int i = 0; i < edge[u].size(); i++){
            int to = edge[u][i].to;
            int cost = edge[u][i].cost; 
            if (d[u]+cost < d[to]){
                d[to] = d[u]+cost;
                if (!vis[to]){
                    vis[to] = true;
                    que.push(to);
                    cnt[to]++;
                    if (cnt[to] > n) return true;    
                }  
            }
        }
    }
    return false;
}

int main() {
    int t;
    int a, b, c;
    
    cin >>t;
    while(t--){
        scanf("%d%d%d", &n, &m, &w);
        for(int i = 1; i <= 500; i++) edge[i].clear();
        for(int i = 1; i <= m; i++){
            scanf("%d%d%d", &a, &b, &c);
            edge[a].push_back(ed(b,c));    
            edge[b].push_back(ed(a,c));    
        }
        for(int i = 1; i <= w; i++){
            scanf("%d%d%d", &a, &b, &c);
            edge[a].push_back(ed(b, -c));
        }
        if (spfa()) printf("YES
");
        else printf("NO
");
    }

    return 0;
}