Description

Seiji Hayashi had been a professor of the Nisshinkan Samurai School in the domain of Aizu for a long time in the 18th century. In order to reward him for his meritorious career in education, Katanobu Matsudaira, the lord of the domain of Aizu, had decided to grant him a rectangular estate within a large field in the Aizu Basin. Although the size (width and height) of the estate was strictly specified by the lord, he was allowed to choose any location for the estate in the field. Inside the field which had also a rectangular shape, many Japanese persimmon trees, whose fruit was one of the famous products of the Aizu region known as 'Mishirazu Persimmon', were planted. Since persimmon was Hayashi's favorite fruit, he wanted to have as many persimmon trees as possible in the estate given by the lord. 
For example, in Figure 1, the entire field is a rectangular grid whose width and height are 10 and 8 respectively. Each asterisk (*) represents a place of a persimmon tree. If the specified width and height of the estate are 4 and 3 respectively, the area surrounded by the solid line contains the most persimmon trees. Similarly, if the estate's width is 6 and its height is 4, the area surrounded by the dashed line has the most, and if the estate's width and height are 3 and 4 respectively, the area surrounded by the dotted line contains the most persimmon trees. Note that the width and height cannot be swapped; the sizes 4 by 3 and 3 by 4 are different, as shown in Figure 1. 
 
Figure 1: Examples of Rectangular Estates
Your task is to find the estate of a given size (width and height) that contains the largest number of persimmon trees.

Input

The input consists of multiple data sets. Each data set is given in the following format.


W H 
x1 y1 
x2 y2 
... 
xN yN 
S T

N is the number of persimmon trees, which is a positive integer less than 500. W and H are the width and the height of the entire field respectively. You can assume that both W and H are positive integers whose values are less than 100. For each i (1 <= i <= N), xi and yi are coordinates of the i-th persimmon tree in the grid. Note that the origin of each coordinate is 1. You can assume that 1 <= xi <= W and 1 <= yi <= H, and no two trees have the same positions. But you should not assume that the persimmon trees are sorted in some order according to their positions. Lastly, S and T are positive integers of the width and height respectively of the estate given by the lord. You can also assume that 1 <= S <= W and 1 <= T <= H.

The end of the input is indicated by a line that solely contains a zero.

Output

For each data set, you are requested to print one line containing the maximum possible number of persimmon trees that can be included in an estate of the given size.

Sample Input

16
10 8
2 2
2 5
2 7
3 3
3 8
4 2
4 5
4 8
6 4
6 7
7 5
7 8
8 1
8 4
9 6
10 3
4 3
8
6 4
1 2
2 1
2 4
3 4
4 2
5 3
6 1
6 2
3 2
0

Sample Output

4
3

【题意】给出一个n*m的土地,圈一个w*h的小块地,要求有最多的trees;

【思路】枚举所有可能的圈地方式,用二维树状数组算出该圈地得到了多少个*,取最大结果输出

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
const int N=;
int c[N][N];//二维树状数组
int n,m;
int lowbit(int x)
{
return x&(-x);
}
void update(int x,int y)
{
for(int i=x;i<=n;i+=lowbit(i))
{
for(int j=y;j<=m;j+=lowbit(j))
{
c[i][j]++;
}
}
}
int get_sum(int x,int y)
{
int ans=;
for(int i=x;i>=;i-=lowbit(i))
for(int j=y;j>=;j-=lowbit(j))
{
ans+=c[i][j];
}
return ans;
}
int main()
{
int t;
while(scanf("%d",&t),t)
{
memset(c,,sizeof(c));
scanf("%d%d",&n,&m);
for(int i=;i<t;i++)
{
int x,y;
scanf("%d%d",&x,&y);
update(x,y); }
int w,h;
scanf("%d%d",&w,&h);
int ans=-;
for(int i=w;i<=n;i++)//枚举
{
for(int j=h;j<=m;j++)
{
int tmp=get_sum(i,j)-get_sum(i,j-h)-get_sum(i-w,j)+get_sum(i-w,j-h);//求子矩形中的tree的数量
if(ans<tmp)
ans=tmp;
}
}
printf("%d\n",ans);
}
return ;
}

Get Many Persimmon Trees_枚举&&二维树状数组的更多相关文章

  1. POJ 2029 Get Many Persimmon Trees (二维树状数组)

    Get Many Persimmon Trees Time Limit:1000MS    Memory Limit:30000KB    64bit IO Format:%I64d & %I ...

  2. POJ 2029 Get Many Persimmon Trees 【 二维树状数组 】

    题意:给出一个h*w的矩形,再给出n个坐标,在这n个坐标种树,再给出一个s*t大小的矩形,问在这个s*t的矩形里面最多能够得到多少棵树 二维的树状数组,求最多能够得到的树的时候,因为h,w都不超过50 ...

  3. POJ2029:Get Many Persimmon Trees(二维树状数组)

    Description Seiji Hayashi had been a professor of the Nisshinkan Samurai School in the domain of Aiz ...

  4. POJ 2029 Get Many Persimmon Trees(DP||二维树状数组)

    题目链接 题意 : 给你每个柿子树的位置,给你已知长宽的矩形,让这个矩形包含最多的柿子树.输出数目 思路 :数据不是很大,暴力一下就行,也可以用二维树状数组来做. #include <stdio ...

  5. POJ 2029 Get Many Persimmon Trees (模板题)【二维树状数组】

    <题目链接> 题目大意: 给你一个H*W的矩阵,再告诉你有n个坐标有点,问你一个w*h的小矩阵最多能够包括多少个点. 解题分析:二维树状数组模板题. #include <cstdio ...

  6. hdu5517 二维树状数组

    题意是给了 n个二元组 m个三元组, 二元组可以和三元组 合并生成3元组,合并条件是<a,b> 与<c,d,e>合并成 <a,c,d> 前提是 b==e, 如果存在 ...

  7. POJ 2029 (二维树状数组)题解

    思路: 大力出奇迹,先用二维树状数组存,然后暴力枚举 算某个矩形区域的值的示意图如下,代码在下面慢慢找... 代码: #include<cstdio> #include<map> ...

  8. HDU 5517 【二维树状数组///三维偏序问题】

    题目链接:[http://acm.split.hdu.edu.cn/showproblem.php?pid=5517] 题意:定义multi_set A<a , d>,B<c , d ...

  9. 【二维树状数组】【CF10D】 LCIS

    传送门 Description 给你两个串,求他们的最长公共上升子序列 Input 第一行是第一个串的长度\(n\) 第二行\(n\)个数代表第一个串 第三行是第二个串的长度\(m\) 第四行\(m\ ...

随机推荐

  1. javascript中IE与ff的区别

    1.自定义属性问题:可以使用获取常规属性的方法来获取自定义属性,也可以使用getAtribute()获取自定义属性,ff下只能使用getAttribute()获取自定义属性. 2. 在IE中可以用ev ...

  2. 《Learning Play! Framework 2》学习笔记——案例研究1(Templating System)

    注解: 这是对<Learning Play! Framework 2>第三章的学习 本章是一个显示聊天记录的项目,只有一个页面,可以自动对聊天记录进行排序.分组和显示,并整合使用了less ...

  3. php -l 检查文件是否语法错误

    有时候在进行网页开发的时候,后台文件的语法错误比较难检查出来,这时候使用php -l filename可对文件的语法进行检查.

  4. ThoughtWorks微服务架构交流心得

      ThoughtWorks微服务架构交流心得: (1)<人月神话>中谈到软件开发没有银弹,根源在于软件所解决的领域问题本身固有的复杂性,微服务正是从领域问题角度上进行服务拆分,来降低软件 ...

  5. 9个充满想象力的 JavaScript 物理和重力实验

    在这个列表中挑选了9个物理和重力实验,用来展示 Javascript 的强大.几年前,所有这些实验都必须使用 Java 或 Flash 才能做.在下面这些惊人的例子中,就个人而言,我比较喜欢仿真布料的 ...

  6. jQuery滚动条回到顶部或指定位置

    jQuery滚动条回到顶部或指定位置 在很多网站,为了增强用户体验,我们会看到回到顶部的按钮,不用手动拖拽滚动条就能回到顶部,非常方便.下面就介绍用jquery实现的滚动到顶部的代码 $(functi ...

  7. InitializeComponent System.StackOverflowException

    因为一直重复调用了InitializeComponent,WPF报System.StackOverflowException错误,提示死循环似的调用.经过一阵排查得出下面结论: 避免在隐藏代码中使用O ...

  8. 在 Linux 中怎样将 MySQL 迁移到 MariaDB 上

    自从甲骨文收购 MySQL 后,由于甲骨文对 MySQL 的开发和维护更多倾向于闭门的立场,很多 MySQL 的开发者和用户放弃了 MySQL.在社区驱动下,促使更多人移到 MySQL 的另一个叫 M ...

  9. css默认样式

    html, address,blockquote,body, dd, div,dl, dt, fieldset, form,frame, frameset,h1, h2, h3, h4,h5, h6, ...

  10. HTML内容整理

    <html> <head> <meta http-equiv="Content-Type" content="text/html; char ...