Poj 3259 Wormholes【spfa 负环判断】

最新推荐文章于 2020-08-07 11:29:12 发布

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最新推荐文章于 2020-08-07 11:29:12 发布

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分类专栏： poj 最短路&&k短路

本文链接：https://blog.youkuaiyun.com/liuke19950717/article/details/51134882

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探讨在一个包含特殊路径（wormholes）的图中寻找是否存在能够回到过去的时间旅行路径的方法。通过SPFA算法检测是否存在负权回路，从而实现理论上的时间倒流。

Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 40566		Accepted: 14864

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

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

USACO 2006 December Gold

题意：

给出n 顶点，以及某些点之间移动需要消耗的时间，另有一些单向的边，权值为负，表示可以时间可以倒流多少，给出整个图的信息，问这个人能否在任意起点出发，最终回到起点时的时间在出发的时间之前

题解：“

不难发现，如果图中出现了边的总权值为负的环，那么肯定可以满足题意（无限在环上跑，时间不是问题），也就是需要进行是否有负环的判断，

用spfa 进行最短路操作的时候，可以顺带发现负环...

/*
http://blog.youkuaiyun.com/liuke19950717
*/
#include<cstdio>
#include<cstring>
#include<queue>
#include<algorithm>
using namespace std;
const int inf=0x3f3f3f3f;
const int maxn=2505;
int edgenum,head[maxn];
struct node
{
	int from,to,val;
	int next;
}edge[maxn*maxn];
void init()
{
	edgenum=0;
	memset(head,-1,sizeof(head));
}
void add(int a,int b,int c)
{
	node tp={a,b,c,head[a]};
	edge[edgenum]=tp;
	head[a]=edgenum++;
}
void slove(int n,int st)
{
	int dist[maxn],vis[maxn]={0},num[maxn]={0};
	memset(dist,inf,sizeof(dist));
	vis[st]=1;num[st]=1;dist[st]=0;
	queue<int> q;
	q.push(st);
	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].val)
			{
				dist[v]=dist[u]+edge[i].val;
				if(!vis[v])
				{
					vis[v]=1;++num[v];
					q.push(v);
					if(num[v]>n)//负环 
					{
						printf("YES\n");
						return;
					}
				}
			}
		}
	}
	printf("NO\n");
}
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		int n,m,w;
		scanf("%d%d%d",&n,&m,&w);
		init();
		for(int i=0;i<m+w;++i)
		{
			int a,b,c;
			scanf("%d%d%d",&a,&b,&c);
			i<m?add(a,b,c),add(b,a,c):add(a,b,-c);
		}
		slove(n,1);
	}
	return 0;
}

ps：

本来该是学最短路的时候做这样的题目，到现在才做，真是够了.....图论也忘记的差不多了....