如何在数组中查找连通分量

Question

我正在尝试 hackerrank 中的算法问题，Roads and Libraries。问题背后的想法是使用 DFS 使用数组查找连通分量 (CC)。

这是测试用例：

queries = [
  {
    n_cities_roads: [9,2],
    c_lib_road: [91, 84],
    matrix: [
      [8, 2], [2, 9]
    ]
  },
  {
    n_cities_roads: [5,9],
    c_lib_road: [92, 23],
    matrix: [
      [2,1], [5, 3], [5,1], 
      [3,4], [3,1],  [5, 4], 
      [4,1], [5,2],  [4,2]
    ]
  },
  {
    n_cities_roads: [8,3],
    c_lib_road: [10, 55],
    matrix: [
      [6,4], [3,2], [7,1]
    ]
  },
  {
    n_cities_roads: [1, 0],
    c_lib_road: [5, 3],
    matrix: []
  },

  {
    n_cities_roads: [2, 0],
    c_lib_road: [102, 1],
    matrix: []
  }
]

queries.each do |query|
  (n_city, n_road), (c_lib, c_road) = [*query[:n_cities_roads]], [*query[:c_lib_road]]
  roads_and_libraries n_city, c_lib, c_road, query[:matrix]
end

输出应该是：

我下面的当前解决方案可以在某些情况下获得 CC，但不是所有情况。

def dfs(i, visited, matrix)
  visited[i] = true
  unless matrix[i].nil?
    matrix[i].each do |j|
      unless visited[j]
        dfs j, visited, matrix
      end
    end
  end
end

def roads_and_libraries(no_cities, c_lib, c_road, cities)
  return c_lib * no_cities if c_lib <= c_road
  visited, count = Array.new(no_cities, false), 0

  (0..no_cities).each do |i|
    unless visited[i]
      count += 1
      dfs i, visited, cities
    end
  end
  p (c_road * (no_cities - count)) + (c_lib * count)
end

我上面的代码测试的输出是：

我正在努力了解如何正确使用 DFS 来查找连通分量。不确定我哪里出错了。

Answer 1

只需打印这一行：

p roads_and_libraries n_city, c_lib, c_road, query[:matrix]

不是这个

p (c_road * (no_cities - count)) + (c_lib * count)

因为方法中有一个return：

return c_lib * no_cities if c_lib <= c_road

不懂算法，不过好像矩阵不能为空[]，至少得[[1,1]]才能得到需要的输出：

roads_and_libraries 1, 5, 3, [[1,1]] #=> 5
roads_and_libraries 2, 102, 1, [[1,1]] #=> 204

因此，要处理空矩阵，一种方法是将其添加为 dfs 方法中的第一行：

matrix = [[1,1]] if matrix.empty?

如何在数组中查找连通分量

How to find connected components in an array

ruby

arrays

algorithm

depth-first-search

connected-components