POJ 2155 Matrix（二维树状数组+数组数组区间更新+单点查询）

Given an N*N matrix A, whose elements are either 0 or 1. A[i, j] means the number in the i-th row and j-th column. Initially we have A[i, j] = 0 (1 <= i, j <= N). 

We can change the matrix in the following way. Given a rectangle whose upper-left corner is (x1, y1) and lower-right corner is (x2, y2), we change all the elements in the rectangle by using "not" operation (if it is a '0' then change it into '1' otherwise change it into '0'). To maintain the information of the matrix, you are asked to write a program to receive and execute two kinds of instructions. 

1. C x1 y1 x2 y2 (1 <= x1 <= x2 <= n, 1 <= y1 <= y2 <= n) changes the matrix by using the rectangle whose upper-left corner is (x1, y1) and lower-right corner is (x2, y2). 
2. Q x y (1 <= x, y <= n) querys A[x, y]. 

Input

The first line of the input is an integer X (X <= 10) representing the number of test cases. The following X blocks each represents a test case. 

The first line of each block contains two numbers N and T (2 <= N <= 1000, 1 <= T <= 50000) representing the size of the matrix and the number of the instructions. The following T lines each represents an instruction having the format "Q x y" or "C x1 y1 x2 y2", which has been described above. 

Output

For each querying output one line, which has an integer representing A[x, y]. 

There is a blank line between every two continuous test cases. 

Sample Input

1
2 10
C 2 1 2 2
Q 2 2
C 2 1 2 1
Q 1 1
C 1 1 2 1
C 1 2 1 2
C 1 1 2 2
Q 1 1
C 1 1 2 1
Q 2 1

Sample Output

1
0
0
1

题解：

题目意思就是给一个大小为n*n的二维矩阵，一开始全部是0，然后操作数是t，如果输出字符为"Q"就是询问矩阵x，y位置的值，为"C"，然后输出两个点坐标，表示矩阵的左上角和右下角坐标，在这些矩阵里面的值取反。。不知道用线段树怎么做，后来看一篇二维树状数组的博客看懂了。

首先写出这题你要会树状数组的区间更新，没错树状数组也可以区间更新，原理不太清楚qaq我就先讲一下做法，就是假如要在一个区间x,y上加上一个值v，那么就在数组sum[x]处加上v，在sum[y+1]处减去v，区间更新就完成了，不同的是当查询单点处x的情况的时候，单点的值是sum[1]+...+sum[x]，累加起来就是单点处的值了。

如何说说我的理解：

在sum[x]处加上v，因为更新是往后更新的（每次加上x&（-x）），所以这时后面的区间每个都多了个v，然后再sum[y+1]处减去v，就相当于把区间外多加上的v去掉，这就完成了[x,y]内加上v，因为这里的区间更新比较特殊，所以查询的时候每一点的值要把前面的区间值全部加上。。。感觉我说了我自己都不太懂，就这样吧

然后因为讲解这题要画图感受，所以我就直接贴我看过的博客好了点击打开链接

这个博客的图对这题的理解很有帮助，二维中，假如图是n*n的，如果在[x,y]处加了v就相当于以左上角坐标[x，y]，到右下角坐标[n*n]的大矩形里面全部值加上了v，然后后续操作就是修剪加多了的地方，加上剪多了的地方最后就完成了更新

代码：

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