Matrix
| Time Limit: 3000MS | Memory Limit: 65536K | |
| Total Submissions: 22058 | Accepted: 8219 |
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 二维树状数组，跟一维的差不多， 这个道题的思路就是看看x1,y1往上加一，同时方块右边，下面和右下方的区域再加1，只要保证他们那边加个偶数就可以了。
1 #include2 #include 3 #include 4 #include 5 using namespace std; 6 const int MAX = 1000 + 5; 7 int c[MAX][MAX]; 8 int n; 9 int lowbit(int k)10 {11 return k & (-k);12 }13 void update(int x,int y,int num)14 {15 for(int i = x; i < n; i += lowbit(i))16 {17 for(int j = y; j < n; j += lowbit(j))18 c[i][j] += num;19 }20 }21 int sum(int x,int y)22 {23 int s = 0;24 for(int i = x; i > 0; i -= lowbit(i))25 {26 for(int j = y; j > 0; j -= lowbit(j))27 s += c[i][j];28 }29 return s;30 }31 int main()32 {33 int t,q;34 int num = 0;35 scanf("%d", &t);36 while(t--)37 {38 if(num ++)39 printf("\n");40 scanf("%d%d", &n,&q);41 memset(c,0,sizeof(c));42 char ch;43 int x1,y1,x2,y2;44 getchar();45 while(q--)46 {47 scanf("%c", &ch);48 if(ch == 'Q')49 {50 scanf("%d%d", &x1,&y1);51 getchar();52 int m = sum(x2,y2) - sum(x1 - 1, y2) - sum(x2,y1 - 1) + sum(x1-1,y1-1);53 if(m % 2 == 0)54 printf("0\n");55 else56 printf("1\n");57 }58 else if(ch == 'C')59 {60 scanf("%d%d%d%d",&x1,&y1,&x2,&y2);61 getchar();62 update(x1,y1,1);63 }64 }65 }66 return 0;67 }