Matrix
| Time Limit: 3000MS | Memory Limit: 65536K | |
| Total Submissions: 17226 | Accepted: 6461 |
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
12 10C 2 1 2 2Q 2 2C 2 1 2 1Q 1 1C 1 1 2 1C 1 2 1 2C 1 1 2 2Q 1 1C 1 1 2 1Q 2 1
Sample Output
1001
Source
,Lou Tiancheng
题意：给出矩阵左上角和右下角坐标，矩阵里的元素 1变0 ，0 变1，然后给出询问，问某个点是多少
思路：二维树状数组
1 //2017-10-25 2 #include3 #include 4 #include 5 #include 6 7 using namespace std; 8 9 const int N = 1100;10 11 int bt[N][N], n, q;12 13 int lowbit(int x){14 return x&(-x);15 }16 17 void add(int x, int y, int v){18 while(x <= n){19 int j = y;20 while(j <= n){21 bt[x][j] += v;22 j += lowbit(j);23 }24 x += lowbit(x);25 }26 }27 28 int sum(int x, int y){29 int sm = 0;30 while(x > 0){31 int j = y;32 while(j > 0){33 sm += bt[x][j];34 j -= lowbit(j);35 }36 x -= lowbit(x);37 }38 return sm;39 }40 41 int main()42 {43 int T;44 cin>>T;45 while(T--){46 scanf("%d%d", &n, &q);47 memset(bt, 0, sizeof(bt));48 char op;49 int x, y, x1, y1;50 while(q--){51 getchar();52 scanf("%c%d%d", &op, &x, &y);53 if(op == 'C'){54 scanf("%d%d", &x1, &y1);55 add(x, y, 1);56 add(x, y1+1, -1);57 add(x1+1, y, -1);58 add(x1+1, y1+1, 1);59 }else{60 printf("%d\n", sum(x, y)%2);61 }62 }if(T)printf("\n");63 }64 65 return 0;66 }