【二分图+最大匹配】北大 poj 2446 Chessboard

 

 

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

	URL   : http://poj.org/problem?id=2446
	Name  : 2446 Chessboard

	Date  : Tuesday, December 6, 2011
	Time Stage : half and one hour

	Result: 
9631289	panyanyany
2446
Accepted	176K	16MS	C++
								

Test Data :

Review :
具体的解释，可以看一看大牛的解题报告：
飘过的小牛:  http://blog.csdn.net/niushuai666/article/details/7031267
小媛在努力~: http://blog.csdn.net/zxy_snow/article/details/6249446
//----------------------------------------------------------------------------*/

#include <stdio.h>
#include <string.h>

#define MAXSIZE		33

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

struct {
	int x, y ;
} link[MAXSIZE][MAXSIZE] ;

bool	cover[MAXSIZE][MAXSIZE], graph[MAXSIZE][MAXSIZE] ;

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

bool find (int py, int px)
{
	int i, j ;
	int x, y ;
	for (i = 0 ; i < 4 ; ++i)
	{
		y = py + dir[i][0] ;
		x = px + dir[i][1] ;
		if ((y >= 0 && y < m && x >= 0 && x < n) &&	// 判断边界
			(cover[y][x] == false)				 && 
			!((y+x) & 1)						 &&	// 横纵坐标和 为偶数
			(map[y][x]))
		{
			cover[y][x] = true ;
			if ((link[y][x].x == -1) || find (link[y][x].y, link[y][x].x))
			{
				link[y][x].y = py ;
				link[y][x].x = px ;
				return true ;
			}
		}
	}
	return false ;
}

int main ()
{
	int i, j ;
	int y, x ;
	int sum ;
	while (~scanf ("%d%d%d", &m, &n, &k))
	{
		memset (map, true, sizeof (map)) ;
		for (i = 0 ; i < k ; ++i)
		{
			scanf ("%d%d", &x, &y) ;
			// 它的默认图坐标是从0开始数的，而实际输入的是从1开始数的……
			map[y-1][x-1] = 0 ;
		}
		
		// 若可用格子数为奇数，则必然不能放满
		if ((m * n - k) & 1)
		{
			puts ("NO") ;
			continue ;
		}
		
		memset (link, -1, sizeof (link)) ;
		sum = 0 ;
		for (i = 0 ; i < m ; ++i)
			for (j = 0 ; j < n ; ++j)
			{
				// 从 横纵坐标和 为奇数的格子开始找
				if (((i+j) & 1) && map[i][j])
				{
					memset (cover, 0, sizeof (cover)) ;
					sum += find (i, j) ;
				}
			}
			
			//		printf ("%d\n", sum) ;
			if (n*m - sum*2 == k)
				puts ("YES") ;
			else
				puts ("NO") ;
	}
	return 0 ;
}