杭电 hdu 1528 Card Game Cheater （二分图，最大匹配）

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/* THE PROGRAM IS MADE BY PYY */
/*----------------------------------------------------------------------------//
Copyright (c) 2011 panyanyany All rights reserved.

URL   : http://acm.hdu.edu.cn/showproblem.php?pid=1528
Name  : 1528 Card Game Cheater

Date  : Saturday, August 27, 2011
Time Stage : two hours

Result:
4502995	2011-08-27 16:50:52	Accepted	1528
0MS	236K	2156 B
C++	pyy


Test Data:


Review:
一开始写得很复杂，其实只要在输入的时候把牌的面值和花色转为整数存在一个整型值
里就行了，这样比较起来方便些。
感觉这个貌似是最大匹配，貌似还是单向的？从 eve 的边集出发，向 adam 的边集前进，
如果 eve 比 adam 大则表示此路可通，否则不可通。求这种通路的最大数量。
这样感觉起来，应该是个单向的最大匹配了吧？小弟不才，对二分图理解甚少，只能这样
猜测了……
//----------------------------------------------------------------------------*/

#include 
#include 
#include 

#define max(a, b) (((a) > (b)) ? (a) : (b))
#define min(a, b) (((a) < (b)) ? (a) : (b))

#define FALSE	0
#define TRUE	1

#define infinity    0x0f0f0f0f
#define minus_inf    0x80808080

#define MAXSIZE	27

int tcase, k ;
int cover[MAXSIZE], link[MAXSIZE], adam[MAXSIZE], eve[MAXSIZE] ;

int makeInt (char c)
{
	int sign = 10, d ;
	if ('0' < c && c <= '9')
		d = c - '0' ;
	else if (c == 'T') d = 10 ;
	else if (c == 'J') d = 11 ;
	else if (c == 'Q') d = 12 ;
	else if (c == 'K') d = 13 ;
	else if (c == 'A') d = 14 ;
	else {
		sign = 1 ;
		if (c == 'C') d = 1 ;
		else if (c == 'D') d = 2 ;
		else if (c == 'S') d = 3 ;
		else if (c == 'H') d = 4 ;
	}
	return d * sign ;
}

int find (int cur)
{
	int i, j ;
	for (i = 0 ; i < k ; ++i)
	{
		if (cover[i] == FALSE && eve[cur] > adam[i])
		{
			cover[i] = TRUE ;
			if (link[i] == -1 || find (link[i]))
			{
				link[i] = cur ;
				return 1 ;
			}
		}
	}
	return 0 ;
}

int main ()
{
	char a, b, c ;
	int i, j, sum ;
	while (scanf ("%d", &tcase) != EOF)
	{
		while (tcase--)
		{
			scanf ("%d", &k) ;
			getchar () ;
			for (i = 0 ; i < k ; ++i)
			{
				scanf ("%c%c", &a, &b) ;
				getchar () ;
				adam[i] = makeInt (a) + makeInt (b) ;
			}
			for (i = 0 ; i < k ; ++i)
			{
				scanf ("%c%c", &a, &b) ;
				getchar () ;
				eve[i] = makeInt (a) + makeInt (b) ;
			}
			
			sum = 0 ;
			memset (link, -1, sizeof (link)) ;
			for (i = 0 ; i < k ; ++i)
			{
				memset (cover, 0, sizeof (cover)) ;
				sum += find (i) ;
			}
			printf ("%d\n", sum) ;
		}
	}
	return 0 ;
}