Topcoder SRM 663 Div2 Hard: CheeseRolling（状压DP）

Problem Statement
	N people (where N is a power of 2) are taking part in a single-elimination tournament in cheese rolling. The diagram below illustrates the structure of the tournament bracket. The people entering the tournament are numbered from 0 to N-1. For each potential cheese rolling match you know who would win the match. You are given this information encoded as a vector wins with N elements, each containing N characters. For each valid i and j, wins[i][j] is 'Y' if person i beats person j. Otherwise, wins[i][j] is 'N'. The relation is not necessarily transitive: it may be the case that person i beats person j, person j beats person k, and person k beats person i. There are N! (N factorial) ways to assign the people to positions in the bracket. Different assignments may produce a different winner of the tournament. Return a vector with N elements. For each valid i, element i of the return value should be the exact number of assignments for which person i wins the tournament.
Definition
	Class: CheeseRolling Method: waysToWin Parameters: vector Returns: vector Method signature: vector waysToWin(vector wins) (be sure your method is public)
Limits
	Time limit (s): 4.000 Memory limit (MB): 256 Stack limit (MB): 256
Constraints
-	N will be between 2 and 16, inclusive.
-	N will be a power of 2.
-	wins will contain exactly N elements.
-	Each element of wins will have a length of exactly N.
-	Each element of wins will be composed of the characters 'Y' and 'N'.
-	For each i from 0 to N-1, wins[i][i] = 'N'.
-	For all distinct integers i and j from 0 to N-1, exactly one of wins[i][j] and wins[j][i] will be 'Y'.
Examples
0)
	{"NN", "YN"} Returns: {0, 2 } There are 2 ways to assign the players: Player 0 goes to position 0 and player 1 goes to position 1. Player 1 goes to position 0 and player 0 goes to position 1. In both assignments, player 1 will win the match against player 0 because wins[1][0] = 'Y'.
1)
	{"NYNY", "NNYN", "YNNY", "NYNN"} Returns: {8, 0, 16, 0 }

大致题意：

<=16个人举办如图所示的淘汰赛，已给出每两个人之间对决的输赢结果矩阵

求对于每个人来说，有多少种安排位置方案使这个人最后获胜

思路：

状压DP，还是有点复杂的状压

对于第x个人最终获胜有多少种方式可以表示为dp[x][(1<

然后递归向下，枚举x能战胜的对手y

然后把集合S分为两半，一半包含x，一半包含y

则有：dp[x][S] += dp[x][S1]*dp[y][S2]*2(相对位置可以变)

枚举半个集合显然就是枚举子集，所以总的复杂度是O(n*3^n)

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