Question

我正在尝试找到地图最短路径的路线（连接的节点有重量/距离）。

我们假设我有这样的哈希：

{"A"=>{"B"=>1, "E"=>1}, "B"=>{"A"=>1, "C"=>1, "F"=>1}, "C"=>{"B"=>1, "D"=>1, "G"=>1}, "D"=>{"C"=>1, "H"=>1}, "E"=>{"F"=>1, "A"=>1, "I"=>1}, "F"=>{"E"=>1, "G"=>1, "B"=>1, "J"=>1}, "G"=>{"F"=>1, "H"=>1, "C"=>1, "K"=>1}, "H"=>{"G"=>1, "D"=>1, "L"=>1}, "I"=>{"J"=>1, "E"=>1, "M"=>1}, "J"=>{"I"=>1, "K"=>1, "F"=>1, "N"=>1}, "K"=>{"J"=>1, "L"=>1, "G"=>1, "O"=>1}, "L"=>{"K"=>1, "H"=>1, "P"=>1}, "M"=>{"N"=>1, "I"=>1}, "N"=>{"M"=>1, "O"=>1, "J"=>1}, "O"=>{"N"=>1, "P"=>1, "K"=>1}, "P"=>{"O"=>1, "L"=>1}}

现在我想遍历并从这个哈希中创建一个路径。例如：

从源A到目的地：L

输出应为：A -> E -> I -> J -> K -> L或A -> B -> C -> D -> H -> L。

这是我写的函数：

def find_path(src, dst, init = [])
  path = [src]
  neighbors = self.neighbors(src)
  puts "src: #{src}"
  puts "dst: #{dst}"
  # puts "node: #{node}"
  puts "init: #{init}"
  puts "path: #{path}"
  puts "----\n"

  if neighbors.include?(dst)
    path.push(dst)
  else
    path.push(@nodes[src].keys.map{|k| k unless init.flatten.include? k }.reject(&:blank?).each{|key| self.find_path(key, dst, init << path) } )
  end
  return path
end

但是，这只打印：[“A”，[“B”，“E”]]

哪个不是所需的输出，有人能告诉我如何才能使这个工作？感谢。

更新：这用于定位地图最短路径的路线（连接的节点具有重量/距离）。以下是我在此问题中尝试实现的详细信息：https://gist.github.com/suryart/6439102

Answer 1

原始哈希：

h = {"A"=>{"B"=>1, "E"=>1}, "B"=>{"A"=>1, "C"=>1, "F"=>1}, "C"=>{"B"=>1, "D"=>1, "G"=>1}, "D"=>{"C"=>1, "H"=>1}, "E"=>{"F"=>1, "A"=>1, "I"=>1}, "F"=>{"E"=>1, "G"=>1, "B"=>1, "J"=>1}, "G"=>{"F"=>1, "H"=>1, "C"=>1, "K"=>1}, "H"=>{"G"=>1, "D"=>1, "L"=>1}, "I"=>{"J"=>1, "E"=>1, "M"=>1}, "J"=>{"I"=>1, "K"=>1, "F"=>1, "N"=>1}, "K"=>{"J"=>1, "L"=>1, "G"=>1, "O"=>1}, "L"=>{"K"=>1, "H"=>1, "P"=>1}, "M"=>{"N"=>1, "I"=>1}, "N"=>{"M"=>1, "O"=>1, "J"=>1}, "O"=>{"N"=>1, "P"=>1, "K"=>1}, "P"=>{"O"=>1, "L"=>1}}

转换原始哈希，因为它的格式很糟糕：

h.keys.each{|k| h[k] = h[k].keys}
h.default = []

方法：

def find_path h, src, dst
  paths = [[src]]
  loop do
    paths = paths.flat_map do |path|
      h[path.last].map do |nekst|
        a = [*path, nekst]
        a.last == dst ? (return a) : a
      end
    end
  end
end

试一试：

find_path(h, "A", "L")
# => ["A", "B", "C", "D", "H", "L"]

请注意，如果没有解决方案，那么循环可能会永远运行。您可能希望通过添加长度限制来限制它。

如何从散列中创建路径？

1 个答案: