2029

/*
起初还想用DP来着,好像是不大行
现在有了树状数组,二维的一下子就解决了,方便,快捷 
*/

// include file
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <cctype>
#include <ctime>

#include <iostream>
#include <sstream>
#include <fstream>
#include <iomanip>
#include <bitset>

#include <algorithm>
#include <string>
#include <vector>
#include <queue>
#include <set>
#include <list>
#include <functional>

using namespace std;

// typedef
typedef long long LL;
typedef unsigned long long ULL;

// 
#define read freopen("in.txt","r",stdin)
#define write freopen("out.txt","w",stdout)
#define FORi(a,b,c) for(int i=(a);i<(b);i+=c)
#define FORj(a,b,c) for(int j=(a);j<(b);j+=c)
#define FORk(a,b,c) for(int k=(a);k<(b);k+=c)
#define FORp(a,b,c) for(int p=(a);p<(b);p+=c)
#define FORii(a,b,c) for(int ii=(a);ii<(b);ii+=c)
#define FORjj(a,b,c) for(int jj=(a);jj<(b);jj+=c)
#define FORkk(a,b,c) for(int kk=(a);kk<(b);kk+=c)

#define FF(i,a)    for(int i=0;i<(a);i++)
#define FFD(i,a)   for(int i=(a)-1;i>=0;i--)

#define Z(a) (a<<1)
#define Y(a) (a>>1)

const double eps = 1e-6;
const double INFf = 1e100;
const int INFi = 1000000000;
const LL INFll = (LL)1<<62;
const double Pi = acos(-1.0);

template<class T> inline T sqr(T a){return a*a;}
template<class T> inline T TMAX(T x,T y)
{
	if(x>y) return x;
	return y;
}
template<class T> inline T TMIN(T x,T y)
{
	if(x<y) return x;
	return y;
}
template<class T> inline void SWAP(T &x,T &y)
{
	T t = x;
	x = y;
	y = t;
}
template<class T> inline T MMAX(T x,T y,T z)
{
	return TMAX(TMAX(x,y),z);
}
template<class T> inline T MMIN(T x,T y,T z)
{
	return TMIN(TMIN(x,y),z);
}


// code begin
int N; // <500
int W,H;
int Bit[110][110];
int Nx,Ny;
int S,T;

inline int lowBit(int x)
{
	return x&(-x);
}

void update(int x,int y,int c)
{
	for(int i=x;i<Nx;i+=lowBit(i))
	{
		for(int j=y;j<Ny;j+=lowBit(j))
		{
			Bit[i][j] += c;
		}
	}
}

int getSum(int x,int y)
{
	int ans = 0;
	for(int i=x;i>0;i-=lowBit(i))
	{
		for(int j=y;j>0;j-=lowBit(j))
		{
			ans += Bit[i][j];
		}
	}
	return ans;
}

int main()
{
	read;
	write;
	int x,y,ans,tmp;
	while(scanf("%d",&N)==1)
	{
		if(!N) break;
		scanf("%d %d",&W,&H);
		Nx = W+1;
		Ny = H+1;
		memset(Bit,0,sizeof(Bit));
		while(N--)
		{
			scanf("%d %d",&x,&y);
			update(x,y,1);
		}
		scanf("%d %d",&S,&T);
		ans = 0;
		FORi(1,W+1,1)
		{
			if(i+S-1>W) break;
			FORj(1,H+1,1)
			{
				if(j+T-1>H) break;
				// i,j  i+S-1,j+T-1
				tmp = getSum(i+S-1,j+T-1) - getSum(i+S-1,j-1)-getSum(i-1,j+T-1)+getSum(i-1,j-1);
				if(tmp>ans) 
					ans = tmp;
			}
		}
		printf("%d\n",ans);
	}
	return 0;
}
posted @ 2011-04-13 16:34  AC2012  阅读(149)  评论(0编辑  收藏  举报