题目链接：～(￣▽￣～)(～￣▽￣)～

资料链接：\(▔▽▔)/

code：

#include <stdio.h>
int n = 0, m = 0, w = 0, count = 0;
typedef struct
{
	int u, v, weight;
}node;
node edge[6000];
int bellman_ford()
{
	int i = 0, j = 0, dis[505], u = 0, v = 0, weight = 0, flag = 0;
	for(i = 0; i<=n; i++)
		dis[i] = 1234567890;
	dis[1] = 0;
	for(i = 0; i<n-1; i++)//循环时循环n-1个点，应为每次最少找出1条最短边，共有n-1条边
	{
		flag = 1;
		for(j = 0; j<count; j++)
		{
			u = edge[j].u;  v = edge[j].v;  weight = edge[j].weight;
			if(dis[v]>dis[u]+weight)
			{
				dis[v] = dis[u]+weight;
				flag = 0;
			}
		}
		if(flag)
			break;
	}
	for(i = 0; i<count; i++)//有负权回路时会无限循环
	{
		u = edge[i].u;  v = edge[i].v;  weight = edge[i].weight;
		if(dis[v]>dis[u]+weight)
			return 1;
	}
	return 0;
}
int main()
{
	int i = 0, t = 0,s = 0, e = 0, text = 0;
	scanf("%d",&text);
	while(text--)
	{
		count = 0;
		scanf("%d %d %d",&n,&m,&w);
		for(i = 0; i<m; i++)
		{
			scanf("%d %d %d",&s, &e, &t);
			edge[count].u = s;
			edge[count].v = e;
			edge[count++].weight = t;
			edge[count].v = s;
			edge[count].u = e;
			edge[count++].weight = t;
		}
		for(i = 0; i<w; i++)
		{
			scanf("%d %d %d",&s,&e,&t);
			edge[count].u = s;
			edge[count].v = e;
			edge[count++].weight = -t;
		}
		if(bellman_ford())
			printf("YES\n");
		else
			printf("NO\n");
	}
	return 0;
}