【poj2155】Matrix(二维树状数组区间更新+单点查询)

Description

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

Sample Output

【题意】

n*n坐标图起初都为0，C：翻转左下和右上两个坐标围成的矩阵中所有点，Q：查询此点的0 1状态

【思路】

二维树状数组区间修改+单点查询模板题，先考虑一维。

一维单点查询就是前缀和，即query(x)。

区间修改先让s-n都加num，再让t+1-n减去num，即update(s, num)，update(t+1, -num)。

二维的单点查询变成二维就好了query(x, y)。

区间修改update(x1, y1, num), update(x2+1, y1, -num), update(x1, y2+1, -num), update(x2+1, y2+1, num)。开始也是黑人问号，在纸上画几笔就看出来了。

这道题问得是0 1状态，那么只要统计翻转次数是偶次还是奇次就行了

【代码】

 #include <iostream>

 #include <cstdio>

 #include <algorithm>

 #include <cstring>

 using namespace std;

 const int N = ;

 int c[N][N], n;

 int lowbit(int x)

 {

     return x&(-x);

 }

 void update(int x, int y, int m)

 {

     for(int i = x; i <= n; i += lowbit(i))

         for(int j = y; j <= n; j += lowbit(j))

             c[i][j] += m;

 }

 int query(int x, int y)

 {

     int sum = ;

     for(int i = x; i > ; i -= lowbit(i))

         for(int j = y; j > ; j-= lowbit(j))

             sum += c[i][j];

     return sum;

 }

 int main()

 {

     int t, m;

     cin>>t;

     while(t--)

     {

         memset(c, , sizeof c);

         scanf("%d%d", &n, &m);

         while(m--)

         {

             char c;

             scanf(" %c", &c);

             if(c == 'C')

             {

                 int x1, x2, y1, y2;

                 scanf("%d%d%d%d", &x1, &y1, &x2, &y2);

                 update(x1, y1, );

                 update(x1, y2+, -);

                 update(x2+, y1, -);

                 update(x2+, y2+, );

             }

             else

             {

                 int x, y;

                 scanf("%d%d", &x, &y);

                 printf("%d\n", query(x, y)&);

             }

         }

         if(t) puts("");

     }

     return ;

 }

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