我希望我没有遇到 NP 难题！
这远非 NP 问题。
我将解释解决问题的两种不同方法。一个将按照您的预期使用Flood Fill，另一个将使用Disjoint-set 数据结构。
洪水填充
假设您有一个矩阵N x M，其中(row, column) 的位置在不使用时为null，否则它包含一个值。
您需要遍历每个列元素1..M 每行1..N。这很简单：
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)
每次找到非null的值都需要调用Flood Fill算法，后面我会定义Flood Fill方法，原因会更清楚。
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
如果您跟踪识别的区域数量，您可以修改此算法以在不同的数组中用指定的数字标记每个区域。您会注意到我使用的是队列而不是递归函数，这将使您远离达到最大递归限制。
并集算法
另一个我认为更昂贵的解决方案是使用不相交集数据结构来完成相同的目的。我只会展示对floodfill 方法的更改。
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
希望对你有所帮助。
因为你不关心它，所以我没有费心展示复杂性。