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Matrix POJ - 2155

程序员文章站 2022-07-12 17:32:28
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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]. 

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. 

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. 

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

二维线段树

AC的C++程序如下:

#include<cstdio>
#include<cstring>
#include<string>
using namespace std;
const int N=1e3+10;
int tree[4*N][4*N];
int n,q;
void updatey(int kx,int y1,int y2,int ly,int ry,int ky)
{
	if(y1<=ly&&y2>=ry)
	{
		tree[kx][ky]=!tree[kx][ky];
		return;
	}
	int mid=(ly+ry)>>1;
	if(y1<=mid) 	updatey(kx,y1,y2,ly,mid,ky<<1);
	if(y2>mid) updatey(kx,y1,y2,mid+1,ry,ky<<1|1);
 } 
 void updatex(int x1,int y1,int x2,int y2,int lx,int rx,int kx)
 {
 	if(x1<=lx&&x2>=rx)
 	{
 		updatey(kx,y1,y2,1,n,1);
 		return;
	 }
	 int mid=(lx+rx)>>1;
	 if(x1<=mid) updatex(x1,y1,x2,y2,lx,mid,kx<<1);
	 if(x2>mid) updatex(x1,y1,x2,y2,mid+1,rx,kx<<1|1);
 }
int qy(int kx,int y,int ly,int ry,int ky)
{
	int cnt=0;
	if(tree[kx][ky]) cnt++;
	if(ly==ry)
	{
		return cnt;
	}
	int mid=(ly+ry)>>1;
	if(y<=mid) cnt+=qy(kx,y,ly,mid,ky<<1);
	else cnt+=qy(kx,y,mid+1,ry,ky<<1|1);
	return cnt;
}
int qx(int x,int y,int lx,int rx,int kx)
{
	int cnt=0;
	cnt+=qy(kx,y,1,n,1);
	if(lx==rx)
	{
		return cnt;
	}
	int mid=(lx+rx)>>1;
	if(x<=mid) cnt+=qx(x,y,lx,mid,kx<<1);
	else cnt+=qx(x,y,mid+1,rx,kx<<1|1);
	return cnt;
}
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d%d",&n,&q);
		memset(tree,0,sizeof(tree));
		while(q--)
		{
			char str[2];
			scanf("%s",str);
			if(str[0]=='C')
			{
				int x1,y1,x2,y2;
				scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
				updatex(x1,y1,x2,y2,1,n,1);
			}
			else
			{
				int x,y;
				scanf("%d%d",&x,&y);
				int ans=qx(x,y,1,n,1);
				printf("%d\n",ans%2);
			}
		}
		if(t) printf("\n");		
	}
	return 0;
}