Matrix(poj2155)

Posted 2020-08-09 SJY

Matrix

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 25139		Accepted: 9314

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

思路：二维树状数组；

http://download.csdn.net/detail/lenleaves/4548401

这个解释的很好；

 1 #include<stdio.h>
 2 #include<algorithm>
 3 #include<iostream>
 4 #include<stdlib.h>
 5 #include<queue>
 6 #include<string.h>
 7 using namespace std;
 8 int bit[1005][1005];
 9 int lowbit(int x)
10 {
11         return x&(-x);
12 }
13 void add(int x,int y)
14 {
15         int i,j;
16         for(i = x; i <= 1000; i+=lowbit(i))
17         {
18                 for(j = y; j <= 1000; j+=lowbit(j))
19                 {
20                         bit[i][j]+=1;
21                         bit[i][j]%=2;
22                 }
23         }
24 }
25 int ask(int x,int y)
26 {
27         int i,j;
28         int sum = 0;
29         for(i = x; i > 0; i-=lowbit(i))
30         {
31                 for(j = y; j > 0; j-=lowbit(j))
32                 {
33                         sum += bit[i][j];
34                 }
35         }
36         return sum%2;
37 }
38 int main(void)
39 {
40         int T;
41         scanf("%d ",&T);
42         while(T--)
43         {
44                 memset(bit,0,sizeof(bit));
45                 int i,j;
46                 int N,q;
47                 scanf("%d %d ",&N,&q);
48                 char a[10];
49                 while(q--)
50                 {
51                         scanf("%s",a);
52                         int x,y,x1,y1;
53                         if(a[0] == ‘C‘)
54                         {
55                                 scanf("%d %d %d %d",&x,&y,&x1,&y1);
56                                 add(x,y);
57                                 add(x1+1,y1+1);
58                                 add(x,y1+1);
59                                 add(x1+1,y);
60                         }
61                         else
62                         {
63                                 scanf("%d %d",&x,&y);
64                                 int ac = ask(x,y);
65                                 printf("%d\n",ac);
66                         }
67                 }
68                 printf("\n");
69         }
70         return 0;
71 }