POJ 2155 Matrix - 二维树状数组/线段树

最新推荐文章于 2020-07-29 17:16:54 发布

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本文介绍如何使用二维树状数组解决特定矩阵操作问题，包括矩阵元素的0/1翻转及查询，通过更新矩阵四角来实现整块区域的变化，并提供了一个具体的编程实现案例。

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.

二维树状数组的变形，也可以使用二维线段树写

需要注意的点是，在0/1变换时，记录变换的次数mod2即可得出答案

此题和树状数组的联系：

每次给矩阵四个角 (x1,y1) (x2+1,y2+1) (x1,y2+1) (x2+1,y1)的变换值+1（利用update），即可更新全部图形上的每个点，具体计算方法即是计算原点到此点所构成的矩阵mod2的值，使用二维树状数组的sum即可

具体证明：对于以下9个区域逐一进行讨论

详见： http://download.youkuaiyun.com/detail/lenleaves/4548401

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

const int maxn=1500;

int c[maxn][maxn];
int n,N,m,q;

inline int lowbit(int x)
{
	return x&(-x);
}
void update(int x,int y)
{
	for(int i=x;i<=n;i+=lowbit(i))
		for(int j=y;j<=n;j+=lowbit(j))
			c[i][j]++;
}
int sum(int x,int y)
{
	int ans=0;
	for(int i=x;i>0;i-=lowbit(i))
		for(int j=y;j>0;j-=lowbit(j))
			ans+=c[i][j];
	return ans;
}
void add()
{
	int x1,y1,x2,y2;
	scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
	getchar();
	update(x2+1,y2+1);
	update(x1,y1);
	update(x1,y2+1);
	update(x2+1,y1);//更新四个角上的点
}
void cal()
{
	int x,y;
	scanf("%d%d",&x,&y);
	getchar();
	printf("%d\n",sum(x,y)%2);//计算矩阵内的总数mod2
}
int main()
{
	scanf("%d",&q);
	for(int k=1;k<=q;k++)
	{
		memset(c,0,sizeof(c));
		scanf("%d%d",&n,&m);
		getchar();
		char order;
		for(int i=1;i<=m;i++)
		{
		    scanf("%c",&order);
			{
				switch(order)
				{
					case 'C':add();break;
					case 'Q':cal();break;
				}
			}
		}
		printf("\n");
	}
}