POJ 题目2155 Matrix(二维树状数组)

Matrix

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].

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

Source

POJ Monthly,Lou Tiancheng

ac代码

#include<stdio.h>
#include<string.h>
int c[1010][1010];
int n;
int lowbit(int x)
{
	return x&(-x);
}
void update(int x,int y)
{
	int i,j;
	for(i=x;i<=n;i+=lowbit(i))
	{
		for(j=y;j<=n;j+=lowbit(j))
		{
			c[i][j]++;
		}
	}
}
int getsum(int x,int y)
{
	int i,j,sum=0;
	for(i=x;i>0;i-=lowbit(i))
	{
		for(j=y;j>0;j-=lowbit(j))
		{
			sum+=c[i][j];
		}
	}
	return sum;
}
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		int m;
		scanf("%d%d",&n,&m);
		char str[2];
		memset(c,0,sizeof(c));
		while(m--)
		{
			scanf("%s",str);
			if(str[0]=='C')
			{
				int x1,y1,x2,y2;
				scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
				x1++,x2++,y1++,y2++;
				update(x2,y2);
				update(x1-1,y1-1);
				update(x1-1,y2);
				update(x2,y1-1);
			}
			else
			{
				int x,y;
				scanf("%d%d",&x,&y);
				printf("%d\n",getsum(x,y)%2);
			}
		}
		if(t)
			printf("\n");
	}
}