Wormholes（题意抽象+SPFA判负权环）_farmer john (also known as fj) has discovered a ma-优快云博客

Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 36128		Accepted: 13198

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

解题思想：这题主要要理解当花费的时间为负时，说明能在起点看到原来的自己。由题意可抽象出判断给出的图中是否存在负权环，存在则输出“YES”，否则输出“NO”。输入的路的边权为正，虫洞的边权为负，而SPFA算法思想在求最短路时，只有负权环对其有影响，也就是说可利用SPFA求最短路的入队出队过程判断是否出现负环。

AC code：

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<queue>
#include<vector>
#include<map>
#include<cmath>
#define LL long long
#define MAXN 1000010
using namespace std;
const int INF=0x3f3f3f3f;
int n,m,tot,k;
int head[MAXN];
struct node{
	int from;
	int to;
	int w;
	int next;
}edge[MAXN];
int dis[MAXN];
bool inq[MAXN];
int cnt[MAXN];
int path[MAXN];
void init()
{
	memset(head,-1,sizeof(head));
	tot=0;
}
void add(int from,int to,int w)
{
	edge[tot].from=from;
	edge[tot].to=to;
	edge[tot].w=w;
	edge[tot].next=head[from];
	head[from]=tot++;
}
bool spfa(int st)
{
	deque<int>q;
	memset(dis,INF,sizeof(dis));
	memset(inq,false,sizeof(inq));
	memset(cnt,0,sizeof(cnt));
	dis[st]=0;
	q.push_back(st);
	inq[st]=true;
	cnt[st]++;
	while(!q.empty())
	{
		int now=q.front();
		q.pop_front();
		inq[now]=false;
		if(cnt[now]>=n)
		{
			return false;//注意：SPFA求最短路时是对存在负权环有影响，因为会一直入队一直循环下去，
		}				//而存在正权环无影响，因为求最短路是递减的过程，正权环不会一直绕下去，但负权环会，
						//而利用SPFA求最长路时正好相反。  
		for(int i=head[now];i!=-1;i=edge[i].next)
		{
			int nex=edge[i].to;
			if(dis[nex]>dis[now]+edge[i].w)
			{
				dis[nex]=dis[now]+edge[i].w;
				if(!inq[nex])
				{
					inq[nex]=true;
					cnt[nex]++;
					if(!q.empty()&&dis[nex]<dis[q.front()])
					{
						q.push_front(nex);
					}
					else
					{
						q.push_back(nex);
					}
				}
			}
		}
	}
	return true;//不存在负权环，返回true 
}
int main()
{
	//freopen("D:\in.txt","r",stdin);
	int T,cas,i,j,u,v,w;
	scanf("%d",&T);
	for(cas=1;cas<=T;cas++)
	{
		scanf("%d%d%d",&n,&m,&k);
		init();
		while(m--)
		{
			scanf("%d%d%d",&u,&v,&w);
			add(u,v,w);
			add(v,u,w);
		}
		while(k--)
		{
			scanf("%d%d%d",&u,&v,&w);
			add(u,v,-w);
		}
		if(spfa(1))
		{
			printf("NO\n");
		}
		else
		{
			printf("YES\n");
		 }  
	}
	return 0;
 }