POJ-2155 Matrix---二维树状数组+区域更新单点查询

题目链接：

https://vjudge.net/problem/POJ-2155

题目大意：

给一个n*n的01矩阵，然后有两种操作（m次）C x1 y1 x2 y2是把这个小矩形内所有数字异或一遍，Q x y 是询问当前这个点的值是多少？n<=1000 m<=50000.

解题思路：

裸的二维树状数组，但是这里是区域更新，单点查询，

做法应该是

  #include<cstdio>
 #include<cstring>
 #include<iostream>
 #include<algorithm>
 #include<string>
 #include<cmath>
 #include<set>
 #include<queue>
 #include<map>
 #include<stack>
 #include<vector>
 #include<list>
 #include<deque>
 #include<sstream>
 #include<cctype>
 #define REP(i, n) for(int i = 0; i < (n); i++)
 #define FOR(i, s, t) for(int i = (s); i < (t); i++)
 using namespace std;
 typedef long long ll;
 typedef unsigned long long ull;
 const int maxn = 1e3 + ;
 const double eps = 1e-;
 const int INF =  << ;
 const int dir[][] = {,,,,,-,-,};
 const double pi = 3.1415926535898;
 int T, n, m, cases;
 int tree[maxn][maxn];
 int lowbit(int x)
 {
     return x&(-x);
 }
 int sum(int x, int y)
 {
     int ret = ;
     for(int i = x; i <= n; i += lowbit(i))
     {
         for(int j = y; j <= n; j += lowbit(j))
             ret += tree[i][j];
     }
     return ret;
 }
 void add(int x, int y, int d)
 {
     for(int i = x; i > ; i -= lowbit(i))
     {
         for(int j = y; j > ; j -= lowbit(j))
             tree[i][j] += d;
     }
 }
 int main()
 {
     std::ios::sync_with_stdio(false);
     cin >> T;
     while(T--)
     {
         cin >> n >> m;
         string c;
         int x1, y1, x2, y2;
         memset(tree, , sizeof(tree));
         while(m--)
         {
             cin >> c;
             if(c[] == 'C')
             {
                 cin >> x1 >> y1 >> x2 >> y2;
                 add(x2, y2, );
                 add(x1 - , y1 - , );
                 add(x2, y1 - , -);
                 add(x1 - , y2, -);
             }
             else if(c[] == 'Q')
             {
                 cin >> x1 >> x2;
                 cout<<(sum(x1, x2)&)<<endl;
             }
         }
         if(T)cout<<endl;
     }
     return ;
 }