Description
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].
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].
Input
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.
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.
Output
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.
There is a blank line between every two continuous test cases.
Sample Input
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
Sample Output
1 0 0
1
这题看了别人的题解，发现做法很奇妙，原打算暴力更新每个点，但发现会超时，所以用了网上的方法，就是对于每个矩形更新4个点，然后最后计算的时候算（1,1）到（x,y)总共翻转的次数%2.
不理解的可以看转载的论文：集训论文
#include<iostream>
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
using namespace std;
#define maxn 1005
int b[maxn][maxn];
char s[10];
int lowbit(int x){
return x&(-x);
}
void update(int x,int y,int num)
{
int i,j;
for(i=x;i<=maxn;i+=lowbit(i)){
for(j=y;j<=maxn;j+=lowbit(j)){
b[i][j]+=num;
}
}
}
int getsum(int x,int y)
{
int num=0,i,j;
for(i=x;i>0;i-=lowbit(i)){
for(j=y;j>0;j-=lowbit(j)){
num+=b[i][j];
}
}
return num;
}
int main()
{
int n,m,i,j,T,x2,x3,y2,y3,x,y;
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&n,&m);
memset(b,0,sizeof(b));
for(i=1;i<=m;i++){
scanf("%s",s);
if(s[0]=='C'){
scanf("%d%d%d%d",&x2,&y2,&x3,&y3);
x2++;y2++;x3++;y3++;
update(x2,y2,1);
update(x3+1,y2,1);
update(x2,y3+1,1);
update(x3+1,y3+1,1);
}
else{
scanf("%d%d",&x,&y);
x++;y++;
printf("%d
",getsum(x,y)%2);
}
}
if(T!=0)printf("
");
}
return 0;
}