poj 3259 Wormholes【spfa判断负环】

Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 36729		Accepted: 13444

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 Tseconds.

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

Hint

For farm 1, FJ cannot travel back in time. 
For farm 2, FJ could travel back in time by the cycle 1->2->3->1, arriving back at his starting location 1 second before he leaves. He could start from anywhere on the cycle to accomplish this.

 

题意：John的农场里有n块地和m条路双向路以及w个虫洞，虫洞是一条单向路，不但会把你传送到目的地，而且时间会倒退T秒。我们的任务是知道会不会在从某块地出发后又回来，看到了离开之前的自己。

说下输入：n块地，m条边，w个虫洞。下面依次是m条边的信息（双向），输入完后是w个虫洞的信息（单向）。

思路：看图中有没有负权环，有的话John可以无限次走这个环，使得时间一定能得到一个负值。所以存在负环话就是可以，没有的话就是不可以了。

#include<stdio.h>
#include<string.h>
#include<queue>
#define MAX 20000
#define INF 0x3f3f3f
using namespace std;
int n,m,ans,s;
int beg,en;
int dis[MAX],vis[MAX];
int head[MAX];
int used[MAX];
struct node
{
	int u,v,w;
	int next;
}edge[MAX];
void init()
{
	ans=0;
	memset(head,-1,sizeof(head));
}
void add(int u,int v,int w)
{
	edge[ans].u=u;
	edge[ans].v=v;
	edge[ans].w=w;
	edge[ans].next=head[u];
	head[u]=ans++;
}
void getmap()
{
    int i,j;
    scanf("%d%d%d",&n,&m,&s);
    while(m--)
    {
    	int a,b,c;
        scanf("%d%d%d",&a,&b,&c);
        add(a,b,c);
        add(b,a,c);
	}
	while(s--)
	{
		int a,b,c;
		scanf("%d%d%d",&a,&b,&c);
		add(a,b,-c);
	}
}
void spfa(int sx)
{
	int i,j;
	bool flag=false;
	queue<int>q;
	memset(vis,0,sizeof(vis));
	memset(used,0,sizeof(used));
	for(i=1;i<=n;i++)
	    dis[i]=INF;
	vis[sx]=1;
	dis[sx]=0;
	used[sx]++;
	q.push(sx);
	while(!q.empty())
	{
		int u=q.front();
		    q.pop();
		vis[u]=0;
		for(i=head[u];i!=-1;i=edge[i].next)
		{
			int top=edge[i].v;
			if(dis[top]>dis[u]+edge[i].w)
			{
				dis[top]=dis[u]+edge[i].w;
				if(!vis[top])
				{
					vis[top]=1;
					q.push(top);
					used[top]++;
					if(used[top]>n)//当一个点进队列大于n次则证明存在负环
					{
						flag=true;
						break;
					}
				}
			}
		}
		if(flag)
		   break;
	}
	if(!flag)
	    printf("NO\n");
	else
	    printf("YES\n");
}
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		init();
		getmap();
		spfa(1);
	}
    return 0;
}