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80 lines (62 loc) · 1.99 KB
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# Program to count islands in boolean 2D matrix
class Graph:
def __init__(self, row, col, g):
self.ROW = row
self.COL = col
self.graph = g
# A function to check if a given cell
# (row, col) can be included in DFS
def isSafe(self, i, j, visited):
# row number is in range, column number
# is in range and value is 1
# and not yet visited
return (i >= 0 and i < self.ROW and
j >= 0 and j < self.COL and
not visited[i][j] and self.graph[i][j])
# A utility function to do DFS for a 2D
# boolean matrix. It only considers
# the 8 neighbours as adjacent vertices
def DFS(self, i, j, visited):
# These arrays are used to get row and
# column numbers of 8 neighbours
# of a given cell
rowNbr = [-1, -1, -1, 0, 0, 1, 1, 1];
colNbr = [-1, 0, 1, -1, 1, -1, 0, 1];
# Mark this cell as visited
visited[i][j] = True
# Recur for all connected neighbours
for k in range(8):
if self.isSafe(i + rowNbr[k], j + colNbr[k], visited):
self.DFS(i + rowNbr[k], j + colNbr[k], visited)
# The main function that returns
# count of islands in a given boolean
# 2D matrix
def countIslands(self):
# Make a bool array to mark visited cells.
# Initially all cells are unvisited
visited = [[False for j in range(self.COL)]for i in range(self.ROW)]
# Initialize count as 0 and travese
# through the all cells of
# given matrix
count = 0
for i in range(self.ROW):
for j in range(self.COL):
# If a cell with value 1 is not visited yet,
# then new island found
if visited[i][j] == False and self.graph[i][j] == 1:
# Visit all cells in this island
# and increment island count
self.DFS(i, j, visited)
count += 1
return count
graph = [[1, 1, 0, 0, 0],
[0, 1, 0, 0, 1],
[1, 0, 0, 1, 1],
[0, 0, 0, 0, 0],
[1, 0, 1, 0, 1]]
row = len(graph)
col = len(graph[0])
g = Graph(row, col, graph)
print "Number of islands is:"
print g.countIslands()
# This code is contributed by Neelam Yadav