http://poj.org/problem?id=2155
Matrix
| Time Limit: 3000MS |
|
Memory Limit: 65536K |
| Total Submissions: 18769 |
|
Accepted: 7078 |
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
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
Source
POJ Monthly,Lou Tiancheng
题意：二维矩阵，可以把子矩阵反转，查询某点的状态。 题解：二维树状数组可搞。 属于树状数组的第二类应用，区间修改单点查值。
（2）“改段求点”型，即对于序列A有以下操作：
【1】修改操作：将A[l..r]之间的全部元素值加上c；
【2】求和操作：求此时A[x]的值。
这个模型中需要设置一个辅助数组B：B[i]表示A[1..i]到目前为止共被整体加了多少（或者可以说成，到目前为止的所有ADD(i, c)操作中c的总和）。
则可以发现，对于之前的所有ADD(x, c)操作，当且仅当x>=i时，该操作会对A[i]的值造成影响（将A[i]加上c），又由于初始A[i]=0，所以有A[i] = B[i..N]之和！而ADD(i, c)（将A[1..i]整体加上c），将B[i]加上c即可――只要对B数组进行操作就行了。
这样就把该模型转化成了“改点求段”型，只是有一点不同的是，SUM(x)不是求B[1..x]的和而是求B[x..N]的和，此时只需把ADD和SUM中的增减次序对调即可（模型1中是ADD加SUM减，这里是ADD减SUM加）。代码：
void ADD(int x, int c)
{
for (int i=x; i>0; i-=i&(-i)) b[i] += c;
}
int SUM(int x)
{
int s = 0;
for (int i=x; i<=n; i+=i&(-i)) s += b[i];
return s;
}
操作【1】：ADD(l-1, -c); ADD(r, c);
操作【2】：SUM(x)。
对于这道题，就是变为二维的而已，取反操作可以看成+1,最后就看是否整除2即可。 比如操作x1,y1,x2,y2的子矩形，就是 add(x2,y2,1);
add(x1-1,y2,-1);
add(x2,y1-1,-1);
add(x1-1,y1-1,1);
画个图想想就知道了。 代码：
/**
* @author neko01
*/
//#pragma comment(linker, "/STACK:102400000,102400000")
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include