杭电 hdu 1507 Uncle Tom's Inherited Land*（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=1507
Name  : 1507 Uncle Tom's Inherited Land*

Date  : Friday，August 27, 2011
Time Stage : 3 hours

Result:
4504524	2011-08-27 21:30:42	Accepted	1507
31MS	420K	2051 B
C++	pyy


Test Data:


Review:
一开始犯了个很白痴的错误，结果做了几个小时……
若将棋盘中相邻的格子分别用黑白染色的话，则同一种颜色的格子必然会出现：
(x+y) 为奇数的情况，所以通过(x+y)&1是否为可以判断是否同一种颜色
//----------------------------------------------------------------------------*/

#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 FALSE	0
#define TRUE	1

#define infinity    0x0f0f0f0f
#define minus_inf    0x80808080

#define MAXSIZE	109

#define SafeAccess(x, y) ((0 < (x) && (x) <= n) && (0 < (y) && (y) <= m))

int n, m, k ;
int map[MAXSIZE][MAXSIZE], cover[MAXSIZE][MAXSIZE] ;

int dir[4][2] = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}} ;

typedef struct tagNODE {
	int x, y ;
} NODE ;

NODE link[MAXSIZE][MAXSIZE] ;


int find (int x, int y)
{
	int i, j, k ;
	for (k = 0 ; k < 4 ; ++k)
	{
		i = x + dir[k][0] ;
		j = y + dir[k][1] ;
		if (SafeAccess (i, j) && !map[i][j] && cover[i][j] == FALSE)
		{
			cover[i][j] = TRUE ;
			if (link[i][j].x == 0 || find (link[i][j].x, link[i][j].y))
			{
				link[i][j].x = x ;
				link[i][j].y = y ;
				return 1 ;
			}
		}
	}
	return 0 ;
}

int main ()
{
	int i, j, x, y, sum ;
	while (scanf ("%d%d", &n, &m), n | m)
	{
		memset (map, 0, sizeof (map)) ;
		scanf ("%d", &k) ;
		for (i = 1 ; i <= k ; ++i)
		{
			scanf ("%d%d", &x, &y) ;
			map[x][y] = 1 ;
		}
		
		sum = 0 ;
		memset (link, FALSE, sizeof (link)) ;

		// 从白色格子开始找
		for (i = 1 ; i <= n ; ++i)
			for (j = 1 ; j <= m ; ++j)
				if (!map[i][j] && (i+j)&1)
				{
					memset (cover, FALSE, sizeof (cover)) ;
					// 找黑色格子的匹配项
					sum += find (i, j) ;
				}
		printf ("%d\n", sum) ;
		for (i = 1 ; i <= n ; ++i)
			for (j = 1 ; j <= m ; ++j)
				if (link[i][j].x)
				{
					printf ("(%d,%d)--(%d,%d)\n", i, j, link[i][j].x, link[i][j].y) ;
				}
	}
	return 0 ;
}