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].
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.
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.
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
题意：
在一个N*N的矩阵里（左上是(1,1)），初始点的值都为0，C(x1,y1,x2,y2)表示将这个矩阵的点都异或，Q(x,y)表示查询点的值（0或者1）。
思路：
常见的二维树状数组是单点更新，区间查询； 而这里是区间更新，单点查询。
由于是单点查询，这里直接用差分的思想做的：a[i][j]表示坐标(i,j)到(n,m)增加多少。
如果矩形(x1,y1,x2,y2)加一，则a[x1][x2]+1;a[x1][y2+1]-1;a[x2][y1+1]-1,a[x2][y2]+1;那么所求点(i,j)的值就是前缀和。
（但如果是区间更新，区间查询，则要像上一题那样推公式，最后得到5个一维树状数组。上一题是3个一维树状数组。）
#include<cstdio> #include<cstdlib> #include<iostream> #include<algorithm> #include<cstring> using namespace std; int n,m,a[1010][1010]; int lowbit(int x){return x&(-x);} void add(int x,int y,int val) { for(int i=x;i<=n;i+=lowbit(i)) for(int j=y;j<=n;j+=lowbit(j)) a[i][j]+=val; } int query(int x,int y) { int res=0; for(int i=x;i;i-=lowbit(i)) for(int j=y;j;j-=lowbit(j)) res+=a[i][j]; return res; } int main() { int T,x1,x2,y1,y2;char opt[3];scanf("%d",&T); while(T--){ scanf("%d%d",&n,&m); memset(a,0,sizeof(a)); for(int i=1;i<=m;i++){ scanf("%s",opt); if(opt[0]=='Q'){ scanf("%d%d",&x1,&y1); printf("%d ",1&(query(x1,y1))); } else { scanf("%d%d%d%d",&x1,&y1,&x2,&y2); add(x1,y1,1);add(x2+1,y2+1,1); add(x2+1,y1,-1);add(x1,y2+1,-1); } } if(T) printf(" "); } return 0; }