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 }