Matrix
| Time Limit: 3000MS | Memory Limit: 65536K | |
| Total Submissions: 16797 | Accepted: 6312 |
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].
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
Source
POJ Monthly,Lou Tiancheng
1 //4100K 438MS C++ 1052B 2014-04-01 21:00:26 2 /* 3 4 题意： 5 给出一个n*n的 二维区间，有小矩形取反操作和对某元素求值操作，对每个操作 6 做出处理。 7 8 树状数组： 9 这题和一般的树状数组不一样，一般的是对某个值更改，对某段求值，现在是刚好 10 反过来，利用反向思维，讲原先更新的操作用来求值， 求值的操作用来更新即可。 11 12 */ 13 #include<stdio.h> 14 #include<string.h> 15 #define N 1005 16 int c[N][N]; 17 int lowbit(int i) 18 { 19 return i&(-i); 20 } 21 int update(int x,int y) 22 { 23 int s=0; 24 for(int i=x;i<N;i+=lowbit(i)) 25 for(int j=y;j<N;j+=lowbit(j)) 26 s+=c[i][j]; 27 return s%2; 28 29 } 30 void getsum(int x,int y) 31 { 32 for(int i=x;i>0;i-=lowbit(i)) 33 for(int j=y;j>0;j-=lowbit(j)) 34 c[i][j]^=1; 35 } 36 int main(void) 37 { 38 int t,n,m; 39 char op; 40 int x1,x2,y1,y2; 41 scanf("%d",&t); 42 while(t--) 43 { 44 scanf("%d%d%*c",&n,&m); 45 memset(c,0,sizeof(c)); 46 while(m--){ 47 scanf("%c",&op); 48 if(op=='Q'){ 49 scanf("%d%d%*c",&x1,&y1); 50 printf("%d ",update(x1,y1)); 51 }else{ 52 scanf("%d%d%d%d%*c",&x1,&y1,&x2,&y2); 53 getsum(x1-1,y1-1); 54 getsum(x1-1,y2); 55 getsum(x2,y1-1); 56 getsum(x2,y2); 57 } 58 } 59 printf(" "); 60 } 61 return 0; 62 }