POJ 3259 Wormholes (最短路 SPFA 判断负环)

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本文介绍了一个有趣的编程问题——POJ3259 Wormholes，该问题涉及在一个包含特殊虫洞的农场网络中寻找是否存在可以返回过去的时间旅行路径。通过使用SPFA算法检测负权回路来解决这一挑战。

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Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 36788		Accepted: 13471

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

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.

注 - 此题为： POJ 3259 Wormholes (最短路 SPFA 判断负环)

题意： N个农场旅行有M个路径和 W个虫洞。有M行普通路径 S ，E， T。表示从地点S(E)到地点E(S)的时间为T。接着有W行为虫洞。普通路径是双向的，虫洞是单向，且为负值

思路：既有虫洞又有普通道路，把虫洞看成是一条负权路，问题就转化成求一个图中是否存在负权回路。定义一个数组记录入队次数，每个元素入队次数是否小于 N 。

已AC代码：

#include<cstdio>
#include<cstring>
#include<queue>
#define MAX 10000
#define INF 0x3f3f3f
using namespace std;

struct Edge{
	int from,to,vel,next;
};
Edge edge[MAX];
int head[MAX];
int dist[MAX],vis[MAX];
int used[MAX];
int N,M,W,cnt;
void addedge(int u,int v,int w)
{
	Edge E={u,v,w,head[u]};
	edge[cnt]=E;
	head[u]=cnt++;
}

int SPFA()
{
	queue<int>Q;
	memset(dist,INF,sizeof(dist));
	memset(vis,0,sizeof(vis));
	memset(used,0,sizeof(used));
	used[1]++;  // 记录元素 入队次数 
	vis[1]=1;  // 起点从 1 开始 
	dist[1]=0;
	Q.push(1);
	while(!Q.empty())
	{
		int u=Q.front();
		Q.pop();
		vis[u]=0;
		for(int i=head[u];i!=-1;i=edge[i].next)
		{
			int v=edge[i].to;
			if(dist[v]>dist[u]+edge[i].vel)
			{
				dist[v]=dist[u]+edge[i].vel;
				if(vis[v]==0)
				{
					used[v]++; // 每次进队 次数加 1  
					if(used[v]>N) // 任意元素入队次数大于 N 说明有负环 
						return 1;
						
					vis[v]=1;
					Q.push(v);
				}
			}
		}
	}
	return 0;
}

int main()
{
	int F,s,e,t,i,j;
	scanf("%d",&F);
	while(F--)
	{
		memset(head,-1,sizeof(head));
		cnt=0;
		scanf("%d%d%d",&N,&M,&W);
		for(i=0;i<M;++i)
		{
			scanf("%d%d%d",&s,&e,&t);
			addedge(s,e,t);
			addedge(e,s,t);
		}
		for(i=0;i<W;++i)  // 虫洞，为单向，负值 
		{
			scanf("%d%d%d",&s,&e,&t);
			addedge(s,e,-t);
		}
		if(SPFA())
			printf("YES\n");
		else
			printf("NO\n");
	}
	return 0;
}