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500，直接暴力

#include<iostream>  
#include<cstdio>  
#include<cstring> 
#include <cmath> 
#include<stack>
#include<vector>
#include<map> 
#include<set>
#include<queue> 
#include<algorithm>  
using namespace std;

typedef long long LL;
const int INF = (1<<30)-1;
const int mod=1000000007;
const int maxn=1005;

int c[maxn][maxn];

int lowbit(int x){ return x & (-x);}

int sum(int x,int y){
    int ret=0;
    int y1;
    while(x > 0){
        y1=y;
        while(y1 > 0){
            ret += c[x][y1];y1-=lowbit(y1);
        }
        x-=lowbit(x);
    }
    return ret;
}

void add(int x,int y,int d){
    int y1;
    while(x < maxn){
        y1=y;
        while(y1 < maxn){
            c[x][y1] += d;y1+=lowbit(y1);
        }
        x+=lowbit(x);
    }
}

int main(){
    int n;
    while(scanf("%d",&n)!=EOF && n){
        memset(c,0,sizeof(c));
        int w,h;
        scanf("%d %d",&w,&h);
        while(n--){
            int u,v;
            scanf("%d %d",&u,&v);
            add(v,u,1);
        }
        int x,y;
        scanf("%d %d",&x,&y);swap(x,y);
        
        int ans=-1;
        int tmp=0;
        for(int i=1;i <=h;i++){
            for(int j=1;j<=w;j++){
                tmp = sum(i+x-1,j+y-1)-sum(i-1,j+y-1)-sum(i+x-1,j-1)+sum(i-1,j-1);
                ans = max(ans,tmp);
            }
        }
        printf("%d
",ans);
    }
    return 0;
}