TOJ 2722 Matrix(树状数组区间取反单点查询)

Posted 2020-12-30 taozi1115402474

描述

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].

输入

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.

输出

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.

样例输入

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

样例输出

1
0
0
1

题意

初始n*n的矩阵全为0

Q个操作

1.[X1,Y1]-[X2,Y2]中取反操作

2.查询[X1,Y1]的值

题解

1.区间更新分成4块，（[X1,Y1]-[n,n]）（[X2,X2]-[n,n]）（[X2+1,Y1]-[n,n]）（[X1,Y2+1]-[n,n]），每个区间都+1操作，只保证[X1,Y1]-[X2,Y2]+1，其余+2或者+4

2.单点查询[X1,Y1]的值，只需要查询[X1,Y1]的值%2即可

代码

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 
 4 const int N=1234;
 5 int n;
 6 
 7 struct BIT2{
 8     int sum[N][N];
 9     void init(){memset(sum,0,sizeof(sum));}
10     int lowbit(int x){return x&(-x);}
11     void update(int x,int y,int w)
12     {
13         for(int i=x;i<=n;i+=lowbit(i))
14             for(int j=y;j<=n;j+=lowbit(j))
15                 sum[i][j]+=w;
16     }
17     int query(int x,int y)
18     {
19         int ans=0;
20         for(int i=x;i>0;i-=lowbit(i))
21             for(int j=y;j>0;j-=lowbit(j))
22                 ans+=sum[i][j];
23         return ans;
24     }
25 }T;
26 
27 int main()
28 {
29     int t,q,o;
30     scanf("%d",&t);
31     while(t--)
32     {
33         if(o++)printf("
");
34         T.init();
35         scanf("%d%d",&n,&q);
36         for(int i=0;i<q;i++)
37         {
38             char op[3];
39             int x1,y1,x2,y2;
40             scanf("%s",op);
41             if(op[0]==‘C‘)
42             {
43                 scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
44                 T.update(x1,y1,1);
45                 T.update(x2+1,y1,1);
46                 T.update(x1,y2+1,1);
47                 T.update(x2+1,y2+1,1);
48             }
49             else
50                 scanf("%d%d",&x1,&y1),printf("%d
",T.query(x1,y1)%2);
51         }
52     }
53 }