题解【POJ2155】Matrix

Description

Given an (N imes 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 leq i, j leq 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 leq x1 leq x2 leq n, 1 leq y1 leq y2 leq 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 leq x, y leq n)) querys (A[x, y]).

Input

The first line of the input is an integer (X (X leq 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 leq N leq 1000, 1 leq T leq 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

Solution

题意简述:一个(N imes N)的(01)矩阵，和几种动态操作，包括对子矩阵((x,y,xx,yy))的所有元素异或，查询某一点((x,y))的元素值。

二维树状数组裸题。

异或的操作不难修改。

二维树状数组与一维树状数组不同的是：

• 一维树状数组维护的是一条链，而二维树状数组维护的却是一片区域。
• 一维树状数组更新和查找只有一重循环，而二维树状数组需要两重循环。
• 二维树状数组比一维树状数组多一维度。

经过以上分析，(AC)代码就不难得出了。

Code

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cctype>

using namespace std;

inline int gi()
{
    int f = 1, x = 0; char c = getchar();
    while (c < '0' || c > '9') { if (c == '-') f = -1; c = getchar();}
    while (c >= '0' && c <= '9') { x = x * 10 + c - '0'; c = getchar();}
    return f * x;
}

int n, m, c[1003][1003], t;
bool fl = false;

inline void add(int x, int y)//二维树状数组的修改操作
{
	for (int i = x; i <= n; i = i + (i & (-i)))
	{
		for (int j = y; j <= n; j = j + (j & (-j)))//注意两重循环
		{
			++c[i][j];
		}
	}
}

inline int getans(int x, int y)//二维树状数组的查询操作
{
	int ans = 0;
	for (int i = x; i; i = i - (i & (-i)))
	{
		for (int j = y; j; j = j - (j & (-j)))
		{
			ans = ans + c[i][j];//加上答案
		}
	}
	return ans;
}

int main()
{
	int t = gi();
	while (t--)
	{
		if (!fl) fl = true;
		else puts("");
		n = gi(), m = gi();
		memset(c, 0, sizeof(c));
		while (m--)
		{
			char c;
			cin >> c;
			if (c == 'C')
			{
				int x = gi(), y = gi(), xx = gi(), yy = gi();
				add(x, y), add(xx + 1, y), add(x, yy + 1), add(xx + 1, yy + 1);//进行插入操作
			}
			else 
			{
				int x = gi(), y = gi();
				printf("%d
", getans(x, y) % 2);//输出最终答案
			}
		}
	}
	return 0;//结束
}