POJ-3660 Cow Contest(Floyd最短路变形)
N (1 ≤ N ≤ 100) cows, conveniently numbered 1..N, are participating in a programming contest. As we all know, some cows code better than others. Each cow has a certain constant skill rating that is unique among the competitors.
The contest is conducted in several head-to-head rounds, each between two cows. If cow A has a greater skill level than cow B (1 ≤ A ≤ N; 1 ≤ B ≤ N; A ≠ B), then cow A will always beat cow B.
Farmer John is trying to rank the cows by skill level. Given a list the results of M(1 ≤ M ≤ 4,500) two-cow rounds, determine the number of cows whose ranks can be precisely determined from the results. It is guaranteed that the results of the rounds will not be contradictory.
Input
* Line 1: Two space-separated integers: N and M
* Lines 2..M+1: Each line contains two space-separated integers that describe the competitors and results (the first integer, A, is the winner) of a single round of competition: A and B
Output
* Line 1: A single integer representing the number of cows whose ranks can be determined
Sample Input
5 5
4 3
4 2
3 2
1 2
2 5
Sample Output
2题意：
一场比赛的数据丢了，这个人给你比赛数据（有向图）让你计算有多少个cow的排名是可以准确得出的。输出准确得出cow的数量。
思路：
- 首先，反向建图，边权值默认初始为1（有连接的，无连接的初始为INF）。如图。
那么这样就行了吗？肯定不行，如果这样去跑一遍Floyd的话，是什么作用没有的。
这里有一个重要的结论（肯定都知道的）：
如果能确定这个点的排名，这个点必然和其他的所有点又有关系，并不一定是直接关系
那这样是不是就知道怎么做了，我们只需要跑一遍Floyd就知道那些点有关系了。NO！！！！
试想一下，这个点如果和另一个有关系，我们只能建立有向图，那么上图2点永远和5点是永远不能有关系的。那么我们不放5->2置位1,2->5置位-1不就好了吗？。最后统计一下i这个点和哪些点有关系(就是两点间的值不是0)就好了。
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