【二分图+简单题】杭电 hdu 1151 Air Raid

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

    URL   : http://acm.hdu.edu.cn/showproblem.php?pid=1151
    Name  : 1151 Air Raid

    Date  : Monday, November 7, 2011
    Time Stage : half an hour

    Result: 
4923333	2011-11-07 20:58:18	Accepted	1151
15MS	316K	1557 B
C++	pyy


Test Data :

Review :
注意是有向图
//----------------------------------------------------------------------------*/

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

using std::vector ;

#define MAXSIZE 129

int		tcase, iStreet, iIntersection ;
int		link[MAXSIZE], cover[MAXSIZE] ;
int		map[MAXSIZE][MAXSIZE] ;

vector<int> adj[MAXSIZE] ;

int find (int cur)
{
	int i, j ;

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

int main ()
{
	int i, j ;
	int s, e ;
	int sum, sumLink ;

	while (~scanf ("%d", &tcase))
	{
		while (tcase--)
		{
			scanf ("%d%d", &iIntersection, &iStreet) ;
			memset (map, 0, sizeof (map)) ;
			for (i = 1 ; i <= iStreet ; ++i)
			{
				scanf ("%d%d", &s, &e) ;
				map[s][e] = 1 ;
//				map[e][s] = 1 ;	若加上这一句，则此图变为无向图
			}

			sum = 0 ;
			memset (link, 0, sizeof (link)) ;
			for (i = 1 ; i <= iIntersection ; ++i)
			{
				memset (cover, 0, sizeof (cover)) ;
				sum += find (i) ;
			}

			sumLink = 0 ;
			for (i = 1 ; i <= iIntersection ; ++i)
			{
				sumLink += link[i] ? 1 : 0 ;
			}

//	由算法的特点可知，若 S, E 两点连通，则 link[S] = 0, link[E] = S ; 因此后面还要减个 sum
//	当然，直接 iIntersection – sumLink 也是一样滴～～
			printf ("%d\n", sum + (iIntersection - sumLink - sum)) ;
		}
	}
	return 0 ;
}