zoj1716简单的二维树状数组

时间:2025-01-31 10:05:20

问一个矩形框在一个大矩形内最多能围几个给定的点

都不用排序,先把所有的点加入树状数组,再直接枚举大矩形的每个格子即可

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <string>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <math.h>
#include <bitset>
#include <list>
#include <algorithm>
#include <climits>
using namespace std; #define lson 2*i
#define rson 2*i+1
#define LS l,mid,lson
#define RS mid+1,r,rson
#define UP(i,x,y) for(i=x;i<=y;i++)
#define DOWN(i,x,y) for(i=x;i>=y;i--)
#define MEM(a,x) memset(a,x,sizeof(a))
#define W(a) while(a)
#define gcd(a,b) __gcd(a,b)
#define LL long long
#define N 1005
#define INF 0x3f3f3f3f
#define EXP 1e-8
#define lowbit(x) (x&-x)
const int mod = 1e9+; int c[N][N],n,m,cnt,s,t; int sum(int x,int y)
{
int ret = ;
int i,j;
for(i = x;i>=;i-=lowbit(i))
{
for(j = y;j>=;j-=lowbit(j))
{
ret+=c[i][j];
}
}
return ret;
} void add(int x,int y,int d)
{
int i,j;
for(i = x;i<=n;i+=lowbit(i))
{
for(j = y;j<=m;j+=lowbit(j))
{
c[i][j]+=d;
}
}
} int main()
{
int i,j,x,y,ans;
while(~scanf("%d",&cnt),cnt)
{
ans = ;
scanf("%d%d",&n,&m);
MEM(c,);
for(i = ;i<=cnt;i++)
{
scanf("%d%d",&x,&y);
add(x,y,);
}
scanf("%d%d",&s,&t);
for(i = ;i<=n;i++)
{
for(j = ;j<=m;j++)
{
int x1 = i,y1 = j,x2 = x1+s-,y2 = y1+t-;
if(x2>n || y2>m) continue;
int s = sum(x2,y2)+sum(x1-,y1-)-sum(x2,y1-)-sum(x1-,y2);
ans = max(ans,s);
}
}
printf("%d\n",ans);
} return ;
}