Time Limit: 3000MS | Memory Limit: 65536K | |
Total Submissions: 33133 | Accepted: 11997 |
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
问题链接:POJ2155 Matrix
问题描述:(略)
问题分析:
这是一个二维树状数组的模板题,不解释。
程序说明:(略)
参考链接:(略)
题记:(略)
AC的C++语言程序如下:
/* POJ2155 Matrix */
#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
const int N = 1000;
int n, a[N + 1][N + 1];
inline int lowbit(int k)
{
return k & -k;
}
void update(int row, int col, int val) {
for (int i = row; i <= n; i += lowbit(i)) {
for (int j = col; j <= n; j += lowbit(j)) {
a[i][j] += val;
}
}
}
int sum(int row, int col) {
int ans = 0;
for (int i = row; i > 0; i -= lowbit(i))
for (int j = col; j > 0; j -= lowbit(j))
ans += a[i][j];
return ans;
}
int main()
{
int x, t;
char cmd[4];
scanf("%d", &x);
while(x--) {
scanf("%d%d", &n, &t);
memset(a, 0, sizeof(a));
while(t--) {
scanf("%s", cmd);
if(cmd[0] == 'C') {
int x1, y1, x2, y2;
scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
update(x1, y1, 1);
update(x2 + 1, y1, 1);
update(x1, y2 + 1, 1);
update(x2 + 1, y2 + 1, 1);
} if(cmd[0] == 'Q') {
int x, y;
scanf("%d%d", &x, &y);
printf("%d\n", sum(x, y) % 2);
}
}
if(x)
printf("\n");
}
return 0;
}