在矩阵中找到具有某些属性的所有矩形区域

问题描述

给出一个n * m矩阵，其可能值为1、2和null：

given an n*m matrix with the possible values of 1, 2 and null:

  . . . . . 1 . .
  . 1 . . . . . 1
  . . . 2 . . . .
  . . . . 2 . . .
  1 . . . . . 1 .
  . . . . . . . .
  . . 1 . . 2 . .
  2 . . . . . . 1

我正在寻找所有块B（包含（x0，y0）和（x1之间的所有值，y1））：

I am looking for all blocks B (containing all values between (x0,y0) and (x1,y1)) that:

• 包含至少一个'1'


• 不包含'2 '


• 不是具有上述属性的另一个块的子集

示例：

红色，绿色和蓝色区域均包含'1'，不包含'2'，并且不属于较大区域。
此图片中当然有3个以上这样的块。我想找到所有这些块。

The red, green and blue area all contain an '1', no '2', and are not part of a larger area.There are of course more than 3 such blocks in this picture. I want to find all these blocks.

什么是找到所有这些区域的快速方法？

what would be a fast way to find all these areas?

我有一个有效的蛮力解决方案，遍历所有可能的矩形，检查它们是否满足前两个条件。然后遍历所有找到的矩形，删除另一个矩形中包含的所有矩形；并且我可以通过先删除连续的相同行和列来加快速度。但我可以肯定，有一种更快的方法。

I have a working brute-force solution, iterating over all possible rectangles, checking if they fulfill the first two criteria; then iterating over all found rectangles, removing all rectangles that are contained in another rectangle; and I can speed that up by first removing consecutive identical rows and columns. But I am fairly certain that there is a much faster way.

推荐答案

我终于找到了一种几乎可以线性运行的解决方案（有一个小因素取决于找到的区域数）。我认为这是最快的解决方案。

I finally found a solution that works almost in linear time (there is a small factor depending on the number of found areas). I think this is the fastest possible solution.

受到此答案的启发：（也从那里拍摄）

Inspired by this answer: https://stackoverflow.com/a/7353193/145999 (pictures also taken from there)

首先，我逐列处理矩阵，创建了一个新矩阵M1测量到最后一个'1'的步数，矩阵M2测量到最后一个'2'的步数

First, I go trought the matrix by column, creating a new matrix M1 measuring the number of steps to the last '1' and a matrix M2 measuring the number of steps to the last '2'

在上图中的任何灰色块中想象一个 1或 2

imagine a '1' or '2' in any of the grey blocks in the above picture

最后，我的M1和M2看起来像这样：

in the end I have M1 and M2 looking like this:

不能按行反向通过M1和M2：

No go through M1 and M2 in reverse, by row:

我执行以下算法：

 foundAreas = new list()

 For each row y backwards:
     potentialAreas = new List()
     for each column x:
        if M2[x,y]>M2[x-1,y]:
            add to potentialAreas:
                new Area(left=x,height=M2[x,y],hasOne=M1[x,y],hasTop=false)
        if M2[x,y]<M2[x-1,y]:
            for each area in potentialAreas:
                 if area.hasTop and area.hasOne<area.height:
                        add to foundAreas:
                             new Box(Area.left,y-area.height,x,y)
            if M2[x,y]==0: delete all potentialAreas
            else:
                 find the area in potentialAreas with height=M2[x,y]
                 or the one with the closest bigger height: set its height to M2[x,y]
                 delete all potentialAreas with a bigger height

            for each area in potentialAreas:
                 if area.hasOne>M1[x,y]: area.hasOne=M1[x,y]
                 if M2[x,y+1]==0: area.hasTop=true

现在找到了具有所需属性的所有矩形。

now foundAreas contains all rectangles with the desired properties.

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