就是运用bellman-ford算法，关键在于理解题意。

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.

大概题意：

John有F个农场，每个农场有若干个点（1<=N<=500），每两个点可以用paths连接（双向，权值为正），也可以用wormholes来连接（单向，权值为负），求是否有一个负权值回路。

#include<iostream>
#include<fstream>

using namespace std;

#define EDGE_MAX 2702
#define TIME_MAX 10001 

int dist[502];

int n, m, w;
int all_edges;

struct EDGE{
	int s;
	int e;
	int t;
}edges[5*EDGE_MAX];

bool fellman_ford()
{
	memset(dist, TIME_MAX, sizeof(dist));
	bool flag = false;
	for (int i = 0; i < n - 1; i++)
	{
		for (int j = 0; j < all_edges; j++)
		{
			if (dist[edges[j].e] > dist[edges[j].s]+edges[j].t)
			{
				dist[edges[j].e] = dist[edges[j].s] + edges[j].t;
				flag = true;
			}
		}
		if (!flag)
			break;
	}
	for (int k = 0; k < all_edges; k++)
	{
		if (dist[edges[k].e]>dist[edges[k].s] + edges[k].t)
			return true;
	}
	return false;
}

int main()
{
	int num,ss,ee,tt;
	//ifstream in("text.txt");
	cin >> num;
	for (int i = 0; i < num; i++)
	{
		cin >> n >> m >> w;
		all_edges = 0;
		for (int j = 0; j < m; j++)
		{
			cin >> ss >> ee >> tt;
			edges[all_edges].s = ss;
			edges[all_edges].e = ee;
			edges[all_edges++].t = tt;
			edges[all_edges].s = ee;
			edges[all_edges].e = ss;
			edges[all_edges++].t = tt;
		}
		for (int k = 0; k < w; k++)
		{
			cin >> ss >> ee >> tt;
			edges[all_edges].s = ss;
			edges[all_edges].e = ee;
			edges[all_edges++].t = -tt;
		}
		if (fellman_ford())
			cout << "YES" << endl;
		else
			cout << "NO" << endl;

	}
	//system("pause");
	return 0;
}