B. Get Many Persimmon Trees

Time Limit: 1000ms
Memory Limit: 30000KB

64-bit integer IO format: %lld      Java class name: Main

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 解题:题目比较长啊。。。开始没高清意思。。。鸟语太挫了。。。给出平面上n个点,最后要在这个平面内找出一个举行内部点数最多的指定长宽的矩形,输出最多包括的点数。
 #include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <climits>
#include <vector>
#include <queue>
#include <cstdlib>
#include <string>
#include <set>
#define LL long long
#define INF 0x3f3f3f
using namespace std;
const int maxn = ;
int tree[maxn][maxn];
int lowbit(int x) {
return x&(-x);
}
void update(int x,int y,int val) {
for(int i = x; i < maxn; i += lowbit(i)) {
for(int j = y; j < maxn; j += lowbit(j)) {
tree[i][j] += val;
}
}
}
int sum(int x,int y) {
int ans = ;
for(int i = x; i; i -= lowbit(i)) {
for(int j = y; j; j -= lowbit(j)) {
ans += tree[i][j];
}
}
return ans;
}
int main() {
int n,w,h,x,y;
while(scanf("%d",&n),n) {
scanf("%d%d",&w,&h);
memset(tree,,sizeof(tree));
for(int i = ; i < n; i++) {
scanf("%d%d",&x,&y);
update(x,y,);
}
scanf("%d%d",&x,&y);
int ans = ,temp;
for(int i = x; i <= w; i++) {
for(int j = y; j <= h; j++) {
temp = sum(i,j) - sum(i-x,j) - sum(i,j-y) + sum(i-x,j-y);
if(temp > ans) ans = temp;
}
}
printf("%d\n",ans);
}
return ;
}

xtu数据结构 B. Get Many Persimmon Trees的更多相关文章

  1. POJ 2029 Get Many Persimmon Trees

    Get Many Persimmon Trees Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 3243 Accepted: 2 ...

  2. POJ2029——Get Many Persimmon Trees

    Get Many Persimmon Trees Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 3656   Accepte ...

  3. (简单) POJ 2029 Get Many Persimmon Trees,暴力。

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

  4. poj2029 Get Many Persimmon Trees

    http://poj.org/problem?id=2029 单点修改 矩阵查询 二维线段树 #include<cstdio> #include<cstring> #inclu ...

  5. POJ-2029 Get Many Persimmon Trees(动态规划)

    Get Many Persimmon Trees Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 3987 Accepted: 2 ...

  6. 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 ...

  7. 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 ...

  8. 数据结构:Binary and other trees(数据结构,算法及应用(C++叙事描述语言)文章8章)

    8.1 Trees -->root,children, parent, siblings, leaf; level, degree of element 的基本概念 8.2 Binary Tre ...

  9. 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 ...

随机推荐

  1. AJPFX简述i=i+1与i+=1及x++的区别和效率

    i=i+1与i+=1及x++的区别和效率 1.x=x+1,x+=1及x++的效率哪个最高?为什么? x=x+1最低,因为它的执行如下. (1)读取右x的地址: (2)x+1: (3)读取左x的地址: ...

  2. 由于js词法性质和全局变量被更改,循环绑定的click事件执行时变量和定义时 不一致的bug,各种解决方案。

    由于js词法性质和全局变量被更改,循环绑定的click事件执行时变量和定义时 不一致的bug,各种解决方案. 动态在页面上添加了5个按钮,实现的功能应该是点击对应按钮在控制台输出相应的索引.但因为应该 ...

  3. Android.mk模板

    此文列出Android.mk的常用模板(部分内容源于多篇他人博客,这里不具体指出),如有错漏,还请在评论中指出,后期持续更新   #链接第三方动态库,在和部分android源码的编译中验证不过 LOC ...

  4. IT之家学院:使用CMD命令行满速下载百度云

    转自:https://www.toutiao.com/a6545305189685920259/?tt_from=android_share&utm_campaign=client_share ...

  5. 在eclipse中查看你用的tomcat的路径

    在eclipse中查看你用的tomcat的路径   打开eclipse,选择window->Preferences->Server->Runtime Environments选择你的 ...

  6. OPENFIRE 使用Hazelcast插件进行集群

    参考资料:http://www.linuxidc.com/Linux/2014-01/94850.htm   https://www.igniterealtime.org/projects/openf ...

  7. iphone开发设置默认字体

    It seems to be possible in iOS 5 using the UIAppearance proxy. [[UILabel appearance] setFont:[UIFont ...

  8. 远程文件拷贝(fastcopy为例)

    远程地址格式如下:\\IP地址\磁盘符号$\文件夹名称(如:127.0.0.1\\c$\\image)拷贝了image文件夹下面的所有文件,但是如果远程机器有密码的话要先在本机先输入远程的目标地址然后 ...

  9. jQuery备忘录

    jquery 中遍历数组 var arr = [1,2,3,4,5] $.each(arr,function(i,j){ console.log(i,j) }) 结果 0 1 1 2 .... jQu ...

  10. MIPS——分支语句

    有关指令 li $t1,immediate #load immediate,立即数可正可负 la $t1,address #load address move $t1,$t2 #move $t2 to ...