POJMatrix(二维树状数组)
2015-11-17 20:57
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Matrix
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
Sample Output
Time Limit: 3000MS | Memory Limit: 65536K | |
Total Submissions: 22058 | Accepted: 8219 |
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
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 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。
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; const int MAX = 1000 + 5; int c[MAX][MAX]; int n; int lowbit(int k) { return k & (-k); } void update(int x,int y,int num) { for(int i = x; i < n; i += lowbit(i)) { for(int j = y; j < n; j += lowbit(j)) c[i][j] += num; } } int sum(int x,int y) { int s = 0; for(int i = x; i > 0; i -= lowbit(i)) { for(int j = y; j > 0; j -= lowbit(j)) s += c[i][j]; } return s; } int main() { int t,q; int num = 0; scanf("%d", &t); while(t--) { if(num ++) printf("\n"); scanf("%d%d", &n,&q); memset(c,0,sizeof(c)); char ch; int x1,y1,x2,y2; getchar(); while(q--) { scanf("%c", &ch); if(ch == 'Q') { scanf("%d%d", &x1,&y1); getchar(); int m = sum(x2,y2) - sum(x1 - 1, y2) - sum(x2,y1 - 1) + sum(x1-1,y1-1); if(m % 2 == 0) printf("0\n"); else printf("1\n"); } else if(ch == 'C') { scanf("%d%d%d%d",&x1,&y1,&x2,&y2); getchar(); update(x1,y1,1); } } } return 0; }
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