poj 3259 Wormholes （负权最短路，SPAF）

Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 36641		Accepted: 13405

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

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.

Source

题目链接：http://poj.org/problem?id=3259

题目大意：时空旅行，前m条路是双向的，旅行时间为正值，w条路是虫洞，单向的，旅行时间是负值，也就是能回到过去。求从一点出发，判断能否在”过去“回到出发点，即会到出发点的时间是负的。

解题思路：裸的负权最短路问题，SPAF解决。枚举出发点即可。

代码如下：

#include 
#include 
#include 
#include 
#include 
#define inf 1e9
using namespace std;
int const maxn=505;
int n,p;
int dist[maxn],vis[maxn],num[maxn];  
struct node   
{ 						
	int v,w;
	node(int vv,int ww)
	{
		v=vv;
		w=ww;
	}
};
vector  vt[maxn];
void SPAF(int v0)
{
	memset(vis,0,sizeof(vis));    
	memset(num,0,sizeof(num));
	for(int i=0;iq;
	q.push(v0);
	dist[v0]=0;            
	while(!q.empty())
	{
		int u=q.front();
		q.pop();
		if(num[u]>n)
			continue;
		num[u]++;
		if(num[u]>=n)
			dist[u]=inf;
		vis[u]=0;
		int len=vt[u].size();         
		for(int i=0;i0)
				{
					p=1;
					return ;
				}
				if(!vis[v])
				{
					vis[v]=1;
					q.push(v);
				}
			}
		}
	}
}
int main(void)
{
	int m1,m2,u,v,w,t;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d%d%d",&n,&m1,&m2);
		int ans=0;
		p=0;
		for(int i=0;i