查找岛屿数量的Python程序 |设置 1（使用 DFS）

给定一个布尔二维矩阵，找出岛屿的数量。一组连接的 1 形成一个岛。例如，下面的矩阵包含 5 个岛屿

例子：

Input : mat[][] = {{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} 
Output : 5

这是标准问题的变体：“计算无向图中的连通分量的数量”。
在我们解决问题之前，让我们了解什么是连接组件。无向图的连通分量是一个子图，其中每两个顶点通过一条路径相互连接，并且不与子图之外的其他顶点相连。
例如，下图具有三个连通分量。

Python

# 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 traverse 
        # 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

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