poj 3259-- Wormholes(SPFA)

                                                                                                           Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 37415		Accepted: 13764

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

Sample Output

NO
YES


#include<stdio.h>
#include<string.h> 
#include<queue>
#include<algorithm>
using namespace std;
#define INF 0x3f3f3f
int m,n,cnt,p;
int head[1010],vis[1010],used[1010],dist[1010];
struct node
{
	int u,v;
	int val,next;
}edge[20010];
void add(int u,int v,int val)
{
	edge[cnt].u=u;
	edge[cnt].v=v;
	edge[cnt].val=val;
	edge[cnt].next=head[u];
	head[u]=cnt++;
}
int SPFA(int st)
{
	queue<int>q;
	memset(dist,INF,sizeof(dist));
	memset(used,0,sizeof(used));
	memset(vis,0,sizeof(vis));
	vis[st]=1;
	dist[st]=0;
	q.push(st);//各种初始化 
	while(!q.empty())
	{
		int u=q.front();
		q.pop();
		vis[u]=0;//进入循环一定要改成0，这样才能判断进入了多少次 
		for(int i=head[u];i!=-1;i=edge[i].next)
		{
			int v=edge[i].v;
			if(dist[v]>dist[u]+edge[i].val)
			{
				dist[v]=dist[u]+edge[i].val;
				if(!vis[v])
				{
					vis[v]=1;
					used[v]++;
					if(used[v]>=m)//判断是否出现了负权边 
					return 1;
					q.push(v);
				}
			}
		}
	}
	return 0;
}
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		memset(head,-1,sizeof(head));
		scanf("%d%d%d",&m,&n,&p);
		int x,y,z;
		while(n--)
		{
			scanf("%d%d%d",&x,&y,&z);
			add(x,y,z);
			add(y,x,z);
		}
		while(p--)
		{
			scanf("%d%d%d",&x,&y,&z);
			add(x,y,-z);
			add(y,x,INF);//虫洞是单向的，所以反向是无穷 
		}
		if(SPFA(1)) printf("YES\n");
		else printf("NO\n");
	}
	return 0;
}