POJ 2029 二维树状数组（400题纪念）

一道二维树状数组的基础题。

还有好多做法。

数据量小都可以A。

贴这题主要是纪念一下POJ 400AC达成~

继续努力！~~~

#include <set>#include <map>#include <stack>#include <cmath>#include <queue>#include <cstdio>#include <string>#include <vector>#include <iomanip>#include <cstring>#include <iostream>#include <algorithm>#define Max 2505#define FI first#define SE second#define ll long long#define PI acos(-1.0)#define inf 0x3fffffff#define LL(x) ( x << 1 )#define bug puts("here")#define PII pair<int,int>#define RR(x) ( x << 1 | 1 )#define mp(a,b) make_pair(a,b)#define mem(a,b) memset(a,b,sizeof(a))#define REP(i,s,t) for( int i = ( s ) ; i <= ( t ) ; ++ i )using namespace std;#define N 200001int n ;int s , e ;int c[555][555] ;int lowbit(int x){return x & (-x) ;}void update(int x , int y ){    for (int i = x ; i <= s ; i += lowbit(i) )        for (int j = y ; j <= e ; j += lowbit(j) )            c[i][j] ++ ;}int query(int x ,int y){    int ans = 0 ;    for (int i = x ; i >= 1 ; i -= lowbit(i))        for (int j = y ; j >= 1 ; j -= lowbit(j))            ans += c[i][j] ;    return ans ;}int query(int x1 ,int y1 ,int x2 ,int y2){    return query(x2 , y2) + query(x1 - 1 , y1 - 1 ) - query(x1 - 1 , y2) - query(x2 , y1 - 1) ;}void AC2029(){    while(cin >> n , n ){        cin >> e >> s ;        mem(c ,0) ;        for (int i = 0 ; i < n ; i ++ ){            int x , y ;            scanf("%d%d",&x,&y) ;            update(y , x) ;        }        int w , h ;        cin >> w >> h ;        int ans = 0 ;        for (int i = 1 ; i <= s ; i ++ ){            for (int j = 1 ; j <= e ; j ++ ){                int x1 = i ;                int y1 = j ;                int x2 = x1 + h - 1 ;                int y2 = y1 + w - 1 ;                if(x2 <= s && y2 <= e)                ans = max(ans , query(x1 , y1 , x2 , y2)) ;            }        }        cout << ans << endl;    }}int main() {    AC2029() ;    return 0 ;}