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

/* THE PROGRAM IS MADE BY PYY *//*----------------------------------------//Copyright (c) 2011 panyanyany All rights reserved.URL   : http://acm.hdu.edu.cn/showproblem.php?pid=1528Name  : 1528 Card Game CheaterDate  : Saturday, August 27, 2011Time Stage : two hoursResult:45029952011-08-27 16:50:52Accepted15280MS236K2156 BC++pyyTest 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 <stdio.h>#include <string.h>#include <stdlib.h>#define max(a, b) (((a) > (b)) ? (a) : (b))#define min(a, b) (((a) < (b)) ? (a) : (b))#define FALSE0#define TRUE1#define infinity    0x0f0f0f0f#define minus_inf    0x80808080#define MAXSIZE27int 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 ;}