Matrix
| Time Limit: 3000MS |
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Memory Limit: 65536K |
| Total Submissions: 20138 |
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Accepted: 7522 |
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
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
题意：poj1566的二维版，有一个矩阵，矩阵中的每个元素只能有0,1，表示，给定一个矩形的左上角和右下角，在这个矩形中的元素进行取反。当输入的字符为Q是，输出地址为x行y列的那个元素是0还是1.
#include
#include
#include
#include
#include
using namespace std; #define maxn 1005 int a[maxn][maxn]; int r,l; int n; inline int Lowbit(int x) { return x & (-x); } int Sum(int x, int y) { int temp, sum = 0; while (x) { temp = y; while (temp) { sum += a[x][temp]; temp -= Lowbit(temp); } x -= Lowbit(x); } return sum; } void Update(int x, int y, int num) { int temp; while (x <= n) { temp = y; while (temp <= n) { a[x][temp] += num; temp += Lowbit(temp); } x += Lowbit(x); } } int main() { int x, m; scanf("%d", &x); while (x--) { memset(a, 0, sizeof(a)); scanf("%d%d", &n, &m); getchar(); l = n; r = n; for (int i = 0; i < m; i++) { char ch; int x1, x2, y2, y1; scanf("%c", &ch); if (ch == 'C') { scanf("%d%d%d%d", &x1, &y1, &x2, &y2); x2++; y2++; Update(x1, y1, 1); Update(x1, y2, -1); Update(x2, y1, -1); Update(x2, y2, 1); } else { scanf("%d%d", &x1, &y1); printf("%d\n", (Sum(x1, y1)%2)); } getchar(); } printf("\n"); } return 0; }