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 1Sample Output
1 0 0 1
给一个n*n矩阵 初始化值全为0。 然后有m个指示进行两个操作 操作1：给出子矩阵的左上角和右下角坐标，矩阵里的元素 1变0 ，0 变1
操作2：给出询问，问当前点(x,y)的值是多少。 这个写得好 http://blog.csdn.net/zxy_snow/article/details/6264135
#include <iostream> #include <algorithm> #include <cstdio> #include <cstring> #include <queue> #include <vector> #include <cmath> #include <stack> #include <string> #include <map> #include <set> #define pi acos(-1) #define LL long long #define ULL unsigned long long #define INF 0x3f3f3f3f #define lson l,mid,rt<<1 #define rson mid+1,r,rt<<1|1 using namespace std; typedef pair<int, int> P; const int maxn = 1e3 + 5; int n, m; int C[maxn<<2][maxn<<2]; int lowbit(int x) { return x & (-x); } void update(int x, int y) { int i, j; for (i = x; i <= n; i += lowbit(i)) for (j = y; j <= n; j += lowbit(j)) C[i][j]++; } int Getsum(int x, int y) { int res = 0, i, j; for (i = x; i != 0; i -= lowbit(i)) for (j = y; j != 0; j -= lowbit(j)) res += C[i][j]; return res; } int main(void) { // freopen("C:\\Users\\wave\\Desktop\\NULL.exe\\NULL\\in.txt","r", stdin); int T, i, j, x1, y1, x2, y2; char s[2]; scanf("%d", &T); while (T--) { memset(C, 0, sizeof(C)); scanf("%d %d", &n, &m); while(m--){ scanf("%s", &s); if (s[0] == 'C'){ scanf("%d %d %d %d", &x1, &y1, &x2, &y2); update(x1, y1); update(x1, y2+1); update(x2+1, y1); update(x2+1, y2+1); } else { scanf("%d %d", &x1, &y1); printf("%d\n", Getsum(x1, y1) % 2); } } printf("\n"); } return 0; }
