POJ 2155 Matrix 二维线段树 区间修改 单点查询

Matrix

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 17423		Accepted: 6530

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

传送门：POJ 2155 Matrix

题目大意：给你一个N*N的矩阵，一开始里面所有的元素值均为0，现在给你m次操作，每次操作是修改或者查询。
当为修改操作时，将矩阵中（x1，y1）到（x2，y2）范围内的元素全部反转（0变1，1变0）。
当为查询操作时，询问点（x，y）在经过之前的操作后的值。

题目分析：该题是线段树的区间修改+单点查询。
由于矩阵是二维的，所以我们用二维线段树即可。
二维线段树就是在一维线段树的基础上，每一个子节点都是一棵线段树（也就是树套树），由此构成。
对于本题，不断的异或就好了。具体看代码。

PS：树状数组很好写并且复杂度超级低

代码如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std ;

#define clear( A , X ) memset ( A , X , sizeof A )
#define lson l , m , o << 1
#define rson m + 1 , r , o << 1 | 1
#define LxRxLyRy Lx , Rx , Ly , Ry

const int maxN = 1005 ;

bool sum[ maxN << 2 ][ maxN << 2 ] ;
int n;

void YUpdate ( int L, int R , int l , int r , int o , int x ) {
	if ( L <= l && r <= R ) sum[ x ][ o ] ^= 1 ;
	else {
		int m = ( l + r ) >> 1 ;
		if ( L <= m ) YUpdate ( L , R , lson , x ) ;
		if ( m <  R ) YUpdate ( L , R , rson , x ) ;
	}
}

void XUpdate ( int Lx , int Rx , int Ly , int Ry , int l , int r , int o ) {
	if ( Lx <= l && r <= Rx ) {
		YUpdate ( Ly , Ry , 1 , n , 1 , o ) ;
	}
	else {
		int m = ( l + r ) >> 1 ;
		if ( Lx <=  m ) XUpdate ( LxRxLyRy , lson ) ;
		if ( m  <  Rx ) XUpdate ( LxRxLyRy , rson ) ;
	}
}
bool YQuery ( int Y , int l , int r , int o , int x ) {
	bool ans = 0 ;
	ans ^= sum[ x ][ o ] ;
	if ( l == r ) return ans;
	int m = ( l + r ) >> 1;
	if ( Y <= m ) ans ^= YQuery ( Y , lson , x ) ;
	else          ans ^= YQuery ( Y , rson , x ) ;
	return ans ;
}

bool XQuery ( int X , int Y , int l , int r , int o ) {
	bool ans = 0 ;
	ans ^= YQuery ( Y , 1 , n , 1 , o ) ;
	if ( l == r ) return ans;
	int m = ( l + r ) >> 1 ;
	if ( X <= m ) ans ^= XQuery ( X , Y , lson ) ;
	else          ans ^= XQuery ( X , Y , rson ) ;
	return ans ;
}

void work ( int m ) {
	clear ( sum , 0 ) ;
	char ch[ 5 ] ;
	int x1 , y1 , x2 , y2 , X , Y;
	while ( m -- ) {
		scanf ( "%s" , ch ) ;
		if ( ch[ 0 ] == 'C' ) {
			scanf ( "%d%d%d%d" , &x1 , &y1 , &x2 , &y2 ) ;
			XUpdate ( x1 , x2 , y1 , y2 , 1 , n , 1 ) ;
		}
		if ( ch[ 0 ] == 'Q' ) {
			scanf ( "%d%d" , &X, &Y ) ;
			printf ( "%d\n" , XQuery ( X , Y , 1 , n , 1 ) ) ;
		}
	}
}

int main () {
	int m , T ;
	for ( scanf ( "%d" , &T ) ; T ; -- T ) {
		scanf ( "%d%d" , &n , &m ) ;
		work ( m ) ;
		if ( T > 1 ) printf ( "\n" ) ;
	}
	return 0 ;
}