POJ-2029 Get Many Persimmon Trees---二维树状数组+枚举

题目链接：

https://vjudge.net/problem/POJ-2029

题目大意：

有N棵树在一个n*m的田里，给出每颗树的坐标

用一个s*t的矩形去围，最多能围几棵树

思路：

用二维树状数组求区域和，也可以直接用二维前缀和求解

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<map>
#include<set>
#include<cmath>
#include<algorithm>
#define lowbot(i) (i&(-i))
using namespace std;
typedef long long ll;
const int maxn = 500 + 10;
int n, m, T, k, cases;
int tree[maxn][maxn];
void add(int x, int y, int d)
{
    for(int i = x; i <= n; i += lowbot(i))
    {
        for(int j = y; j <= m; j += lowbot(j))
        {
            tree[i][j] += d;
        }
    }
}
int sum(int x, int y)
{
    int ans = 0;
    for(int i = x; i > 0; i -= lowbot(i))
    {
        for(int j = y; j > 0; j -= lowbot(j))
        {
            ans += tree[i][j];
        }
    }
    return ans;
}
int Sum(int x1, int y1, int x2, int y2)
{
    return sum(x2, y2) + sum(x1 - 1, y1 - 1) - sum(x1 - 1, y2) - sum(x2, y1 - 1);
}
int main()
{
    while(cin >> k && k)
    {
        int x, y, s, t;
        memset(tree, 0, sizeof(tree));
        cin >> n >> m;
        while(k--)
        {
            scanf("%d%d", &x, &y);
            add(x, y, 1);
        }
        cin >> s >> t;
        int ans = 0;
        for(int i = 1; i + s - 1 <= n; i++)
        {
            for(int j = 1; j + t - 1 <= m; j++)
            {
                int tot = Sum(i, j, i + s - 1, j + t - 1);
                ans = max(ans, tot);
            }
        }
        cout<<ans<<endl;
    }
    return 0;
}