#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <vector>
using namespace std;
/*
* 单源最短路bellman_ford算法，复杂度O(VE)
* 可以处理负边权图。
* 可以判断是否存在负环回路。返回true,当且仅当图中不包含从源点可达的负权回路
* vector<Edge>E;先E.clear()初始化，然后加入所有边
* 点的编号从1开始
*/
const int INF = 0x3f3f3f3f;
const int MAXN = 550;
int dist[MAXN];
struct Edge
{int u, v;int cost;Edge(int _u = 0, int _v = 0, int _cost = 0) :u(_u), v(_v), cost(_cost){}
};
vector<Edge> E;
bool bellman_ford(int start, int n)//点的编号从1开始
{for (int i = 1; i <= n; i++)dist[i] = INF;dist[start] = 0;for (int i = 1; i<n; i++)//最多做n-1次{bool flag = false;for (int j = 0; j<E.size(); j++){int u = E[j].u;int v = E[j].v;int cost = E[j].cost;if (dist[v]>dist[u] + cost){dist[v] = dist[u] + cost;flag = true;}}if (!flag)return true;//没有负环回路}for (int j = 0; j<E.size(); j++)if (dist[E[j].v]>dist[E[j].u] + E[j].cost)return false;//第n次更新则有负环回路return true;//没有负环回路
}int main()
{int T;int N, M, W;int a, b, c;scanf("%d", &T);while (T--){scanf("%d%d%d", &N, &M, &W);E.clear();while (M--){scanf("%d%d%d", &a, &b, &c);E.push_back(Edge(a, b, c));E.push_back(Edge(b, a, c));}while (W--){scanf("%d%d%d", &a, &b, &c);E.push_back(Edge(a, b, -c));}for (int i = 1; i <= N; i++)E.push_back(Edge(N + 1, i, 0));if (!bellman_ford(N + 1, N + 1))printf("YES\n");else printf("NO\n");}return 0;
}

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <vector>
using namespace std;
/*
* 单源最短路bellman_ford算法，复杂度O(VE)
* 可以处理负边权图。
* vector<Edge>E;先E.clear()初始化，然后加入所有边
* 点的编号从1开始
*/
const int INF = 0x3f3f3f3f;
const int MAXN = 550;
int dist[MAXN];
struct Edge
{int u, v;int cost;Edge(int _u = 0, int _v = 0, int _cost = 0) :u(_u), v(_v), cost(_cost){}
};
vector<Edge> E;
bool bellman_ford(int n)//点的编号从1开始
{for (int i = 1; i <= n; i++)dist[i] = 0;for (int i = 1; i<n; i++)//最多做n-1次{bool flag = false;for (int j = 0; j<E.size(); j++){int u = E[j].u;int v = E[j].v;int cost = E[j].cost;if (dist[v]>dist[u] + cost){dist[v] = dist[u] + cost;flag = true;}}if (!flag)return true;//没有负环回路}for (int j = 0; j<E.size(); j++)if (dist[E[j].v]>dist[E[j].u] + E[j].cost)return false;//第n次更新则有负环回路return true;//没有负环回路
}int main()
{int T;int N, M, W;int a, b, c;scanf("%d", &T);while (T--){scanf("%d%d%d", &N, &M, &W);E.clear();while (M--){scanf("%d%d%d", &a, &b, &c);E.push_back(Edge(a, b, c));E.push_back(Edge(b, a, c));}while (W--){scanf("%d%d%d", &a, &b, &c);E.push_back(Edge(a, b, -c));}if (!bellman_ford(N))printf("YES\n");else printf("NO\n");}return 0;
}

#include <iostream>
#include <string.h>
#include <stdio.h>
#include <algorithm>
#include <vector>
#include <queue>
using namespace std;
/*
* 单源最短路SPFA
*/
const int MAXN = 1010;
const int INF = 0x3f3f3f3f;
struct Edge
{int v;int cost;Edge(int _v = 0, int _cost = 0) :v(_v), cost(_cost){}
};vector<Edge> E[MAXN];void addedge(int u, int v, int w)
{E[u].push_back(Edge(v, w));
}bool vis[MAXN];
int cnt[MAXN];
int dist[MAXN];bool SPFA(int start, int n)
{memset(vis, false, sizeof(vis));for (int i = 1; i <= n; i++)dist[i] = INF;dist[start] = 0;vis[start] = true;queue<int>que;while (!que.empty())que.pop();que.push(start);memset(cnt, 0, sizeof(cnt));cnt[start] = 1;while (!que.empty()){int u = que.front();que.pop();vis[u] = false;for (int i = 0; i<E[u].size(); i++){int v = E[u][i].v;if (dist[v]>dist[u] + E[u][i].cost){dist[v] = dist[u] + E[u][i].cost;if (!vis[v]){vis[v] = true;que.push(v);if (++cnt[v]>n)return false;//有负环回路}}}}return true;
}
int main()
{int T;int N, M, W;int a, b, c;scanf("%d", &T);while (T--){scanf("%d%d%d", &N, &M, &W);for (int i = 1; i <= N + 1; i++)E[i].clear();while (M--){scanf("%d%d%d", &a, &b, &c);addedge(a, b, c);addedge(b, a, c);}while (W--){scanf("%d%d%d", &a, &b, &c);addedge(a, b, -c);}for (int i = 1; i <= N; i++)addedge(N + 1, i, 0);if (!SPFA(N + 1, N + 1))printf("YES\n");else printf("NO\n");}return 0;
}