【二分图+有难度】杭电 hdu 1281 棋盘游戏_C/C++_编程开发_程序员俱乐部

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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=1281
    Name  : 1281 棋盘游戏

    Date  : Tuesday, November 8, 2011
    Time Stage : an hour

    Result: 
4930281	2011-11-08 20:57:47	Accepted	1281
31MS	248K	1675 B
C++	pyy


Test Data :

Review :
一开始想不明白，怎么样才能求那些“重要点”，甚至还想到从反面下手，先求
“不重要点”。但还是不得其解，于是看了人家的解题报告，发现可以试着删除
某个点，如果删除后最大的“车”数量有所减少，则证明它是“重要点”。
//----------------------------------------------------------------------------*/

#include <stdio.h>
#include <vector>

using std::vector ;

#define MAXSIZE 109

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

int find (int cur)
{
	int i, j ;
	for (i = 1 ; i <= m ; ++i)
	{
		if (cover[i] == false && map[cur][i] == true)
		{
			cover[i] = true ;
			if (link[i] == 0 || find (link[i]))
			{
				link[i] = cur ;
				return 1 ;
			}
		}
	}
	return 0 ;
}

int getSum ()
{
	int i ;
	int sum ;
	sum = 0 ;
	memset (link, 0, sizeof (link)) ;
	for (i = 1 ; i <= n ; ++i)
	{
		memset (cover, 0, sizeof (cover)) ;
		sum += find (i) ;
	}
	return sum ;
}

int main ()
{
	int i, j ;
	int x, y ;
	int sum, tmpSum, important, tcase ;
	tcase = 0 ;
	while (~scanf ("%d%d%d", &n, &m, &k))
	{
		memset (map, false, sizeof (map)) ;
		for (i = 0 ; i < k ; ++i)
		{
			scanf ("%d%d", &x, &y) ;
			map[x][y] = true ;
		}

		sum = getSum () ;

		// 枚举所有可落“车”的点，分别删除，若车的最大数减少，则证明
		// 该点为 “重要点”
		important = 0 ;
		for (i = 1 ; i <= n ; ++i)
		{
			for (j = 1 ; j <= m ; ++j)
			{
				if (map[i][j])
				{
					map[i][j] = false ;
					tmpSum = getSum () ;
					map[i][j] = true ;

					if (tmpSum < sum)
						++important ;
				}
			}
		}
		printf ("Board %d have %d important blanks for %d chessmen.\n", ++tcase,
			important, sum) ;
	}
	return 0 ;
}