POJ - 2029

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. 

N 
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

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主要是为了练习二维线段树。点修改，区域查询。

用二维线段树写这个题目真的很蠢，随便一个方法都比它好。

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1 #include
2 #include
3 #include
4 #include
5 #include
6
7 using namespace std;
8 const int maxn=101;
9 struct LIE
10 {
11 int ll,lr,sum;
12 };
13 struct HANG
14 {
15 int hl,hr;
16 LIE lie[maxn<<2]; 17 }hang[maxn<<2]; 18 int t; 19 int n,m,w,h,ans=0; 20 void readint(int &x) 21 { 22 char c=getchar(); 23 int f=1; 24 for(;c<'0' || c>'9';c=getchar())if(c=='-')f=-f;
25 x=0;
26 for(;c<='9'&& c>='0';c=getchar())x=(x<<1)+(x<<3)+c-'0'; 27 x*=f; 28 } 29 void writeint(int x) 30 { 31 char s[20]; 32 int js=0; 33 if(!x) 34 { 35 s[0]='0'; 36 js=1; 37 } 38 else 39 { 40 while(x) 41 { 42 s[js]=x%10+'0'; 43 js++;x/=10; 44 } 45 } 46 js--; 47 while(js>=0)putchar(s[js--]);
48 putchar('\n');
49 }
50 void buil(int pre,int cur,int ll,int lr)
51 {
52 hang[pre].lie[cur].ll=ll;hang[pre].lie[cur].lr=lr;
53 hang[pre].lie[cur].sum=0;
54 if(ll==lr)return ;
55 int mid=(ll+lr)>>1;
56 buil(pre,cur<<1,ll,mid); 57 buil(pre,cur<<1|1,mid+1,lr); 58 } 59 void build(int cur,int hl,int hr,int ll,int lr) 60 { 61 hang[cur].hl=hl;hang[cur].hr=hr; 62 buil(cur,1,ll,lr); 63 if(hl==hr)return ; 64 int mid=(hl+hr)>>1;
65 build(cur<<1,hl,mid,ll,lr); 66 build(cur<<1|1,mid+1,hr,ll,lr); 67 } 68 void upda(int pre,int cur,int y) 69 { 70 hang[pre].lie[cur].sum++; 71 if(hang[pre].lie[cur].ll==hang[pre].lie[cur].lr)return; 72 int mid=(hang[pre].lie[cur].ll+hang[pre].lie[cur].lr)>>1;
73 if(y<=mid)upda(pre,cur<<1,y); 74 else upda(pre,cur<<1|1,y); 75 } 76 void update(int cur,int x,int y) 77 { 78 upda(cur,1,y); 79 if(hang[cur].hl==hang[cur].hr)return; 80 int mid=(hang[cur].hl+hang[cur].hr)>>1;
81 if(x<=mid)update(cur<<1,x,y); 82 else update(cur<<1|1,x,y); 83 } 84 int quer(int pre,int cur,int yl,int yr) 85 { 86 if(yl<=hang[pre].lie[cur].ll && hang[pre].lie[cur].lr<=yr)return hang[pre].lie[cur].sum; 87 int mid=(hang[pre].lie[cur].ll+hang[pre].lie[cur].lr)>>1;
88 int ans=0;
89 if(yl<=mid)ans+=quer(pre,cur<<1,yl,yr); 90 if(mid>1;
97 int ans=0;
98 if(xl<=mid)ans+=query(cur<<1,xl,xr,yl,yr); 99 if(xr>mid)ans+=query(cur<<1|1,xl,xr,yl,yr);
100 return ans;
101 }
102 int main()
103 {
104 readint(t);
105 while(t)
106 {
107 readint(n);readint(m);
108 build(1,1,n,1,m);
109 for(int x,y,i=0;i<t;i++)
110 {
111 readint(x);readint(y);
112 update(1,x,y);
113 }
114 readint(w);readint(h);
115 ans=0;
116 for(int i=1;i<=n-w+1;i++)
117 for(int j=1;j<=m-h+1;j++)
118 {
119 ans=max(ans,query(1,i,i+w-1,j,j+h-1));
120 }
121 writeint(ans);
122 readint(t);
123 }
124 return 0;
125 }