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>
#include<iostream>
#include<algorithm> using namespace std; int w,h;
int x,y;
int xl,yl,xr,yr; #define N 101 int sum[N<<][N<<]; int cnt; void read(int &x)
{
x=; char c=getchar();
while(!isdigit(c)) c=getchar();
while(isdigit(c)) { x=x*+c-''; c=getchar(); }
} void changey(int kx,int ky,int l,int r)
{
if(l==r)
{
sum[kx][ky]++;
return;
}
int mid=l+r>>;
if(y<=mid) changey(kx,ky<<,l,mid);
else changey(kx,ky<<|,mid+,r);
sum[kx][ky]=sum[kx][ky<<]+sum[kx][ky<<|];
} void changex(int kx,int l,int r)
{
changey(kx,,,h);
if(l==r) return;
int mid=l+r>>;
if(x<=mid) changex(kx<<,l,mid);
else changex(kx<<|,mid+,r);
} void queryy(int kx,int ky,int l,int r)
{
if(l>=yl && r<=yr)
{
cnt+=sum[kx][ky];
return;
}
int mid=l+r>>;
if(yl<=mid) queryy(kx,ky<<,l,mid);
if(yr>mid) queryy(kx,ky<<|,mid+,r);
} void queryx(int kx,int l,int r)
{
if(l>=xl && r<=xr)
{
queryy(kx,,,h);
return;
}
int mid=l+r>>;
if(xl<=mid) queryx(kx<<,l,mid);
if(xr>mid) queryx(kx<<|,mid+,r);
} int main()
{
int n,s,t;
int ans;
while(scanf("%d",&n)!=EOF)
{
if(!n) return ;
memset(sum,,sizeof(sum));
read(w); read(h);
while(n--)
{
read(x); read(y);
changex(,,w);
}
read(s); read(t);
ans=;
for(int j=t;j<=h;++j)
for(int i=s;i<=w;++i)
{
xl=i-s+;
yl=j-t+;
xr=i;
yr=j;
cnt=;
queryx(,,w);
ans=max(ans,cnt);
}
cout<<ans<<'\n';
}
}
| Time Limit: 1000MS | Memory Limit: 30000K | |
| Total Submissions: 4649 | Accepted: 3025 |
Description
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
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
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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