Algorithm - search for groups of connected plates

I am terribly sorry for the name, but a description of my problem with a few words is a little hard. I thought the rest of the post would explain this better!;)

Description

I basically have a 2D array of fragments / objects / characters, and I want to split it into two (or more) new 2D arrays whenever a group of fragments is separated by special fragments.

For example, if I have:

[x] [x] [0] [0]

[0] [0] [x] [0]

[0] [0] [x] [0]

[0] [0] [0] [x]

If x is not needed then this should give me two new arrays:

[x] [x] [0] [0]

[x] [x] [x] [0]

[x] [x] [x] [0]

[x] [x] [x] [x]

and

[x] [x] [x] [x]

[0] [0] [x] [x]

[0] [0] [x] [x]

[0] [0] [0] [x]

One array for each group of interconnected plates.

x . , B A, , .

, , , , O (n ^ 2), , . , , , , , , .

, , / . , . , , NP- !: 3

!

+3
2

, NP-hard!

NP.

. Flood Fill, , Disjoint-set .

Flood Fill

, N x M, (row, column) null, , .

1..M 1..N. :

for row in range(1, N + 1):
  for column in range(1, M + 1):
    if matrix[row][column] is not null:
      floodfill(matrix, row, column)

Flood Fill , null, Flood Fill .

def floodfill(matrix, row, column):
  # I will use a queue to keep record of the positions we are gonna traverse.
  # Each element in the queue is a coordinate position (row,column) of an element
  # of the matrix.
  Q = Queue()

  # A container for the up, down, left and right directions.
  dirs = { (-1, 0), (1, 0), (0, -1), (0, 1) }

  # Now we will add our initial position to the queue.
  Q.push( (row, column) )

  # And we will mark the element as null. You will definitely need to
  # use a boolean matrix to mark visited elements. In this case I will simply
  # mark them as null.
  matrix[row][column] = null

  # Go through each element in the queue, while there are still elements to visit.
  while Q is not empty:

    # Pop the next element to visit from the queue.
    # Remember this is a (row, column) position.
    (r, c) = Q.pop()

    # Add the element to the output region.
    region.add( (r, c) )

    # Check for non-visited position adjacent to this (r,c) position.
    # These are:
    #   (r + 1, c): down
    #   (r - 1, c): up
    #   (r, c - 1): left
    #   (r, c + 1): right
    for (dr, dc) in dirs:

      # Check if this adjacent position is not null and keep it between
      # the matrix size.
      if matrix[r + dr][c + dc] is not null
         and r + dr <= rows(matrix)
         and c + dc <= colums(matrix):

        # Then add the position to the queue to be visited later
        Q.push(r + dr, c + dc)

        # And mark this position as visited.
        matrix[r + dr][c + dc] = null

  # When there are no more positions to visit. You can return the
  # region visited.
  return region

, , . , , .

, , . floodfill.

def floodfill(matrix):
  disjoint_set = DisjointSet()

  # Go through each row in the matrix
  for row in range(1, N + 1):

    # Go through each column in the matrix
    for column in range(1, M + 1):

      # Create a set for the current position
      disjoint_set.makeSet(row, column)

      if matrix[row - 1][column] is not null:
        # If the position north of it it is not null then merge them
        disjoint_set.merge((row, column), (row - 1, column))

      if matrix[row][column - 1] is not null:
        # If the position left of it it is not null then merge them
        disjoint_set.merge((row, column), (row, column - 1))

  # You can go through each position identifying its set and do something with it
  for row in range(1, N + 1):
    for column in range(1, M + 1):
      regions[ disjoint_set.find(row, column) ] = (row, column)

  return regions

, .

, , .

+6

,

+3

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