spfa 判断负环poj3259

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

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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.

题目大意：给出一个混合图（无向图+有向图），无向图事路，有向图是虫洞，虫洞可以回到过去；

如果能回到过去（经过虫洞后原点时间减少了）能则输出yes，否则输出no

大佬们有好多fang方法判断负环，但是我目前只会一种QAQ。。

spfa模板判断fu'h负环，

首先输入该地图，然后从一个点开始查找最短路，之后就是套用spfa模板了，有负环的存在返回1，无返回0；

对于模板，分为三个步骤

初始化：首先初始化数组shuaxin（用来记录最短路）为INF，biaoji（用于标记是否在队列）为0（不在），num（记录某一点进入队列的次数）为0；

起点入列：

循环判断：取出队首元素，取消标记（不在队列），利用邻接矩阵刷新最短距离，对于没有入列但可以入列de的的点，入列，标记，判断是否该点重复使用n次，ru'g如果重复使用，说明有fu'h负环，返回1；如果走完全程仍没有重复次数>n的，则代表代表没有负环。

#include<stdio.h>
#include<string.h>  
#include<math.h>  
  
//#include<map>   
//#include<set>
#include<deque>  
#include<queue>  
#include<stack>  
#include<bitset> 
#include<string>  
#include<iostream>  
#include<algorithm>  
using namespace std;  
  
#define ll long long  
#define INF 0x3f3f3f3f  
#define clean(a,b) memset(a,b,sizeof(a))// 水印 
#define mod 1000000


int shuzu[510][510];
int shuaxin[510],num[510];
bool biaoji[510];
int n,m,w;

int spfa(int x)
{
	clean(biaoji,0);
	clean(num,0);
	clean(shuaxin,INF);
	queue<int> que;
	while(que.size())
		que.pop();
	//初始化完成
	biaoji[x]=1;
	shuaxin[x]=0;
	num[x]=1;
	que.push(x);
	//起点入列 
	while(que.size())
	{
		int can=que.front();//以改点为起点判断 刷新 
		que.pop();
		biaoji[can]=0;
		for(int i=1;i<=n;++i)
		{
			if(shuaxin[can]+shuzu[can][i]<shuaxin[i])
			{
				shuaxin[i]=shuzu[can][i]+shuaxin[can];//刷新 
				if(biaoji[i]==0)//判断 改点是否在队列 
				{
					biaoji[i]=1;
					num[i]++;//重复++ 
					if(num[i]>n)
						return 1;//有负环 
					que.push(i);
				}
				
			}
		}
	}
	return 0;
	
}

int main()
{
	int t;
	cin>>t;
	while(t--)
	{
		clean(shuzu,INF);
		
		cin>>n>>m>>w;
		for(int i=0;i<m;++i)
		{
			int s,e,l;
			cin>>s>>e>>l;
			if(shuzu[s][e]>l)
			{
				shuzu[s][e]=l;
				shuzu[e][s]=l;
			}
		}
		for(int i=0;i<w;++i)
		{
			int s,e,l;
			cin>>s>>e>>l;
			shuzu[s][e]=-l;//虫洞 
		}
		if(spfa(1))
			cout<<"YES"<<endl;
		else
			cout<<"NO"<<endl;
	}
}

spfa 判断负环poj3259

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