Wormholes POJ 3259【SPFA】

http://poj.org/problem?id=3259DescriptionWhile 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.InputLine 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.OutputLines 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

/*题目大意：虫洞问题，现在有n个点，m条边，代表现在可以走的通路，比如从a到b和从b到a需要花费c时间，现在在地上出现了w个虫洞，虫洞的意义就是你从a到b花费的时间是-c(时间倒流,并且虫洞是单向的)，现在问你从某个点开始走，能回到从前其实就是判断是否存在负环 */#include <cstdio>#include <cstring>#include <queue>#define MAXN 550#define MAXM 10000#define INF 0x3f3f3f3fusing namespace std;struct Edge{	int u, v, w;	int next;//下一个结构体变量的下标 }edge[MAXM];int head[MAXN];//下标为起点u，值为对应结构体下标 int vis[MAXN];//判断是否加入队列了 int used[MAXN];int num;int N;int low[MAXN];//存最短路径 void Add_Edge(int u, int v, int w)//加边 {	Edge E={u, v, w, head[u]};//初始化结构体 	edge[num]=E;//直接赋值 	head[u]=num++;}bool SPFA(int s){	int i, j;	queue<int> Q;	memset(low, INF, sizeof(low)); 	memset(vis, 0, sizeof(vis));	memset(used,0,sizeof(used));		vis[s] = 1;	low[s]=0; 	Q.push(s);	used[s]++;	while(!Q.empty())	{		int u=Q.front();		Q.pop();		vis[u]=0;//出队列了，不在队列就变成0 		for(j = head[u]; j != -1; j = edge[j].next)		{			int v = edge[j].v;			if(low[v] > low[u] + edge[j].w)			{				low[v] = low[u] + edge[j].w;				if(!vis[v] )				 			 	{			 		vis[v]=1;			 		Q.push(v);			 		used[v]++;			 		if(used[v]>N) return 0;			 	}			}		}	}	return 1;}int main(){	int u, v, w;	int T;	int M, W;	scanf("%d",&T);	while(T--)	{		scanf("%d%d%d", &N, &M, &W);		memset(head, -1, sizeof(head));		num=0;		while(M--)		{			scanf("%d%d%d", &u, &v, &w);			Add_Edge(u, v, w);			Add_Edge(v, u, w);//无向边 		}			while(W--)		{			scanf("%d%d%d", &u, &v, &w);			Add_Edge(u, v, -w);		}			if(!SPFA(1)) printf("YES\n");		else printf("NO\n");	}	return 0;}