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


题解：构图，判断图中有没有负环即可，bellman-ford和spfa都可以。坑的是路径是双向的，虫洞是单向的。。。

#include 
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#include 
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#include 
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#include 
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#include 
using namespace std;

const int INF = 0x3f3f3f3f;
const int maxn = 510;

struct edge
{
    int v, d;
};

vector<edge> G[maxn];
int vis[maxn], cnt[maxn], d[maxn];
int n, m, w;

bool spfa(int s)
{
    queue<int> q;
    memset(vis, 0, sizeof(vis));
    memset(cnt, 0, sizeof(cnt));
    fill(d, d+n, INF);
    d[s] = 0;
    q.push(s);
    while (!q.empty())
    {
        int u = q.front();
        q.pop();
        vis[u] = 0;
        for (int i = 0; i < G[u].size(); ++i)
        {
            int v = G[u][i].v, dis = G[u][i].d;
            if (d[v] > d[u] + dis)
            {
                d[v] = d[u] + dis;
                if (!vis[v])
                {
                    q.push(v);
                    vis[v] = 1;
                    if (++cnt[v] > n)
                        return false;
                }
            }
        }
    }
    return true;
}

int main()
{
    int T;
    cin >> T;
    while (T--)
    {
        cin >> n >> m >> w;
        for (int i = 0; i < n; ++i)
            G[i].clear();
        while (m--)
        {
            int u, v, d;
            scanf("%d%d%d", &u, &v, &d);
            G[u-1].push_back({v-1, d});
            G[v-1].push_back({u-1, d});
        }
        while (w--)
        {
            int u, v, d;
            scanf("%d%d%d", &u, &v, &d);
            G[u-1].push_back({v-1, -d});
        }
        if (spfa(0))
            printf("NO\n");
        else
            printf("YES\n");
    }
    return 0;
}