【poj2155】Matrix(二维树状数组区间更新+单点查询)

【poj2155】Matrix(二维树状数组区间更新+单点查询)

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

 

【题意】

n*n坐标图起初都为0，C：翻转左下和右上两个坐标围成的矩阵中所有点，Q：查询此点的0 1状态

【思路】

二维树状数组区间修改+单点查询模板题，先考虑一维。

一维单点查询就是前缀和，即query(x)。

区间修改先让s-n都加num，再让t+1-n减去num，即update(s, num)，update(t+1, -num)。

二维的单点查询变成二维就好了query(x, y)。

区间修改update(x1, y1, num), update(x2+1, y1, -num), update(x1, y2+1, -num), update(x2+1, y2+1, num)。开始也是黑人问号，在纸上画几笔就看出来了。

这道题问得是0 1状态，那么只要统计翻转次数是偶次还是奇次就行了

【代码】

 1 #include 
 2 #include 
 3 #include 
 4 #include 
 5 using namespace std;
 6 const int N = 1005;
 7 int c[N][N], n;
 8 int lowbit(int x)
 9 {
10     return x&(-x);
11 }
12 void update(int x, int y, int m)
13 {
14     for(int i = x; i <= n; i += lowbit(i))
15         for(int j = y; j <= n; j += lowbit(j))
16             c[i][j] += m;
17 }
18 int query(int x, int y)
19 {
20     int sum = 0;
21     for(int i = x; i > 0; i -= lowbit(i))
22         for(int j = y; j > 0; j-= lowbit(j))
23             sum += c[i][j];
24     return sum;
25 }
26 int main()
27 {
28     int t, m;
29     cin>>t;
30     while(t--)
31     {
32         memset(c, 0, sizeof c);
33         scanf("%d%d", &n, &m);
34         while(m--)
35         {
36             char c;
37             scanf(" %c", &c);
38             if(c == 'C')
39             {
40                 int x1, x2, y1, y2;
41                 scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
42                 update(x1, y1, 1);
43                 update(x1, y2+1, -1);
44                 update(x2+1, y1, -1);
45                 update(x2+1, y2+1, 1);
46             }
47             else
48             {
49                 int x, y;
50                 scanf("%d%d", &x, &y);
51                 printf("%d\n", query(x, y)&1);
52             }
53         }
54         if(t) puts("");
55     }
56     return 0;
57 }

 

posted @ 2018-02-27 17:03 LesRoad 阅读(...) 评论(...) 编辑 收藏