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
#include<bits/stdc++.h>
using namespace std; const int N=;
int n; struct BIT2{
int sum[N][N];
void init(){memset(sum,,sizeof(sum));}
int lowbit(int x){return x&(-x);}
void update(int x,int y,int w)
{
for(int i=x;i<=n;i+=lowbit(i))
for(int j=y;j<=n;j+=lowbit(j))
sum[i][j]+=w;
}
int query(int x,int y)
{
int ans=;
for(int i=x;i>;i-=lowbit(i))
for(int j=y;j>;j-=lowbit(j))
ans+=sum[i][j];
return ans;
}
}T; int main()
{
int t,q,o;
scanf("%d",&t);
while(t--)
{
if(o++)printf("\n");
T.init();
scanf("%d%d",&n,&q);
for(int i=;i<q;i++)
{
char op[];
int x1,y1,x2,y2;
scanf("%s",op);
if(op[]=='C')
{
scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
T.update(x1,y1,);
T.update(x2+,y1,);
T.update(x1,y2+,);
T.update(x2+,y2+,);
}
else
scanf("%d%d",&x1,&y1),printf("%d\n",T.query(x1,y1)%);
}
}
}