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
题解:
二维树状数组
对于翻转操作:在二维bit中记录翻转的次数,由于是后缀操作,所以要把不在翻转范围内的区间翻回来
对于询问操作:每次查询(x,y)在bit中的前缀和最后%2就是结果
1 /* 2 挑战2-树状数组 3 二维bit 4 写起来简单,想起来难= = 5 by-solution 6 */ 7 #include<cstdio> 8 #include<cstdlib> 9 #include<cstring> 10 #include<iostream> 11 #include<cmath> 12 #include<algorithm> 13 #define ll long long 14 #define lowbit(x) x&(-x) 15 using namespace std; 16 17 const int N = 1010; 18 19 int n,qu,c[N][N]; 20 21 int gi() { 22 int x=0,o=1; char ch=getchar(); 23 while(ch!='-' && (ch<'0'||ch>'9')) ch=getchar(); 24 if(ch=='-') ch=getchar(),o=-1; 25 while(ch>='0' && ch<='9') x=x*10+ch-'0',ch=getchar(); 26 return o*x; 27 } 28 29 void add(int i, int j) { 30 for(int p=i; p<=n; p+=lowbit(p)) 31 for(int q=j; q<=n; q+=lowbit(q)) 32 c[p][q]++; 33 } 34 35 int query(int i, int j) { 36 int ret=0; 37 for(int p=i; p; p-=lowbit(p)) 38 for(int q=j; q; q-=lowbit(q)) 39 ret+=c[p][q]; 40 return ret; 41 } 42 43 int main() {//poj2155 44 freopen("1.in","r",stdin); 45 freopen("1.out","w",stdout); 46 int T=gi(); 47 while(T--) { 48 memset(c,0,sizeof(c)); 49 n=gi(),qu=gi(); 50 for(int i=1; i<=qu; i++) { 51 char ch; 52 scanf("%c", &ch); 53 if(ch=='C') { 54 int x=gi(),y=gi(),xx=gi(),yy=gi(); 55 add(x,y),add(xx+1,y),add(x,yy+1),add(xx+1,yy+1); 56 } 57 else { 58 int x=gi(),y=gi(); 59 printf("%d ", query(x,y)%2); 60 } 61 } 62 printf(" "); 63 } 64 return 0; 65 }