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【POJ】2536 Gopher II 二分匹配

2014-08-03 15:07 465 查看
传送门:【POJ】2536 Gopher II

题目分析:二分匹配模板题

代码如下:

#include <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std ;

#define REP( i , a , b ) for ( int i = a ; i < b ; ++ i )
#define REV( i , a , b ) for ( int i = a - 1 ; i >= b ; -- i )
#define FOR( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define FOV( i , a , b ) for ( int i = a ; i >= b ; -- i )
#define CLR( a , x ) memset ( a , x , sizeof a )

const int MAXN = 1000 ;
const int MAXQ = 2000 ;
const int MAXE = 100005 ;
const int INF = 0x3f3f3f3f ;

struct Edge {
	int v , n ;
	Edge () {}
	Edge ( int v , int n ) : v ( v ) , n ( n ) {}
} ;

Edge E[MAXE] ;
int H[MAXN] , cntE ;
int Q[MAXQ] , head , tail ;
int Lx[MAXN] , Ly[MAXN] ;
int dx[MAXN] , dy[MAXN] ;
bool vis[MAXN] ;
int n , m , s , v ;
int dis ;
double X[MAXN] , Y[MAXN] ;

void init () {
	cntE = 0 ;
	CLR ( H , -1 ) ;
}

void addedge ( int u , int v ) {
	E[cntE] = Edge ( v , H[u] ) ;
	H[u] = cntE ++ ;
}

int Hopcroft_Krap () {
	dis = INF ;
	CLR ( dx , -1 ) ;
	CLR ( dy , -1 ) ;
	head = tail = 0 ;
	FOR ( i , 1 , n )
		if ( Lx[i] == -1 ) {
			dx[i] = 0 ;
			Q[tail ++] = i ;
		}
	while ( head != tail ) {
		int u = Q[head ++] ;
		if ( dx[u] >= dis )
			continue ;
		for ( int i = H[u] ; ~i ; i = E[i].n ) {
			int v = E[i].v ;
			if ( dy[v] == -1 ) {
				dy[v] = dx[u] + 1 ;
				if ( Ly[v] == -1 )
					dis = dy[v] ;
				else {
					dx[Ly[v]] = dy[v] + 1 ;
					Q[tail ++] = Ly[v] ;
				}
			}
		}
	}
	return dis != INF ;
}

int find ( int u ) {
	for ( int i = H[u] ; ~i ; i = E[i].n ) {
		int v = E[i].v ;
		if ( !vis[v] && dy[v] == dx[u] + 1 ) {
			vis[v] = 1 ;
			if ( ~Ly[v] && dy[v] == dis )
				continue ;
			if ( Ly[v] == -1 || find ( Ly[v] ) ) {
				Lx[u] = v ;
				Ly[v] = u ;
				return 1 ;
			}
		}
	}
	return 0 ;
}

int match () {
	int ans = 0 ;
	CLR ( Lx , -1 ) ;
	CLR ( Ly , -1 ) ;
	while ( Hopcroft_Krap () ) {
		CLR ( vis , 0 ) ;
		FOR ( i , 1 , n )
			if ( Lx[i] == -1 )
				ans += find ( i ) ;
	}
	return ans ;
}

double dist ( double x , double y ) {
	return sqrt ( x * x + y * y ) ;
}

void solve () {
	double x , y ;
	init () ;
	FOR ( i , 1 , n )
		scanf ( "%lf%lf" , &X[i] , &Y[i] ) ;
	FOR ( i , 1 , m ) {
		scanf ( "%lf%lf" , &x , &y ) ;
		FOR ( j , 1 , n )
			if ( v * s >= dist ( X[j] - x , Y[j] - y ) )
				addedge ( j , i ) ;
	}
	printf ( "%d\n" , n - match () ) ;
}
				
int main () {
	while ( ~scanf ( "%d%d%d%d" , &n , &m , &s , &v ) )
		solve () ;
	return 0 ;
}
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