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 ;

 }