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 =  + ;
 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 = ;
     for(int i = x; i > ; i -= lowbot(i))
     {
         for(int j = y; j > ; 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 - , y1 - ) - sum(x1 - , y2) - sum(x2, y1 - );
 }
 int main()
 {
     while(cin >> k && k)
     {
         int x, y, s, t;
         memset(tree, , sizeof(tree));
         cin >> n >> m;
         while(k--)
         {
             scanf("%d%d", &x, &y);
             add(x, y, );
         }
         cin >> s >> t;
         int ans = ;
         for(int i = ; i + s -  <= n; i++)
         {
             for(int j = ; j + t -  <= m; j++)
             {
                 int tot = Sum(i, j, i + s - , j + t - );
                 ans = max(ans, tot);
             }
         }
         cout<<ans<<endl;
     }
     return ;
 }