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 + 10;
const double eps = 1e-10;
const int INF = 1 << 30;
const int dir[4][2] = {1,0,0,1,0,-1,-1,0};
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 = 0;
    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 > 0; i -= lowbit(i))
    {
        for(int j = y; j > 0; 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, 0, sizeof(tree));
        while(m--)
        {
            cin >> c;
            if(c[0] == 'C')
            {
                cin >> x1 >> y1 >> x2 >> y2;
                add(x2, y2, 1);
                add(x1 - 1, y1 - 1, 1);
                add(x2, y1 - 1, -1);
                add(x1 - 1, y2, -1);
            }
            else if(c[0] == 'Q')
            {
                cin >> x1 >> x2;
                cout<<(sum(x1, x2)&1)<<endl;
            }
        }
        if(T)cout<<endl;
    }
    return 0;
}