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 1 #include<bits/stdc++.h> 2 2 using namespace std; 3 3 4 4 const int N=1234; 5 5 int n; 6 6 7 7 struct BIT2{ 8 8 int sum[N][N]; 9 9 void init(){memset(sum,0,sizeof(sum));} 1010 int lowbit(int x){return x&(-x);} 1111 void update(int x,int y,int w) 1212 { 1313 for(int i=x;i<=n;i+=lowbit(i)) 1414 for(int j=y;j<=n;j+=lowbit(j)) 1515 sum[i][j]+=w; 1616 } 1717 int query(int x,int y) 1818 { 1919 int ans=0; 2020 for(int i=x;i>0;i-=lowbit(i)) 2121 for(int j=y;j>0;j-=lowbit(j)) 2222 ans+=sum[i][j]; 2323 return ans; 2424 } 2525 }T; 2626 2727 int main() 2828 { 2929 int t,q,o; 3030 scanf("%d",&t); 3131 while(t--) 3232 { 3333 if(o++)printf("\n"); 3434 T.init(); 3535 scanf("%d%d",&n,&q); 3636 for(int i=0;i<q;i++) 3737 { 3838 char op[3]; 3939 int x1,y1,x2,y2; 4040 scanf("%s",op); 4141 if(op[0]=='C') 4242 { 4343 scanf("%d%d%d%d",&x1,&y1,&x2,&y2); 4444 T.update(x1,y1,1); 4545 T.update(x2+1,y1,1); 4646 T.update(x1,y2+1,1); 4747 T.update(x2+1,y2+1,1); 4848 } 4949 else 5050 scanf("%d%d",&x1,&y1),printf("%d\n",T.query(x1,y1)%2); 5151 } 5252 } 5353 }