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("\n");
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\n",T.query(x1,y1)%2);
51 }
52 }
53 }