spfa 判断负环poj3259
题目链接：http://poj.org/problem?id=3259
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 8Sample Output
NO
YESHint
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。
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//#include 后来又碰见这道题了，用邻接表写了一下，一开始甚至习惯性的想用优先队列。。。果然太年轻，spfa是用先进先出的队列，对每个点进行松弛操作后，判断队列中没有该元素，放入队列，若一个点重复放入n次以上，则说明有负环，而对于优先队列来说，它是按照一定的优先级出队的，因此一个点可能没有影响到队列中的其他点（实际上它应该要影响到）（应该大概解释的清楚了吧。。）ac：
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