我正在尝试实现将有向无环图转换为树的算法(为了好玩、学习、套路、命名)。所以我想出了数据结构节点:

/// <summary>
/// Represeting a node in DAG or Tree
/// </summary>
/// <typeparam name="T">Value of the node</typeparam>
public class Node<T> 
{
    /// <summary>
    /// creats a node with no child nodes
    /// </summary>
    /// <param name="value">Value of the node</param>
    public Node(T value)
    {
        Value = value;
        ChildNodes = new List<Node<T>>();
    }

    /// <summary>
    /// Creates a node with given value and copy the collection of child nodes
    /// </summary>
    /// <param name="value">value of the node</param>
    /// <param name="childNodes">collection of child nodes</param>
    public Node(T value, IEnumerable<Node<T>> childNodes)
    {
        if (childNodes == null)
        {
            throw new ArgumentNullException("childNodes");
        }
        ChildNodes = new List<Node<T>>(childNodes);
        Value = value;
    }

    /// <summary>
    /// Determines if the node has any child node
    /// </summary>
    /// <returns>true if has any</returns>
    public bool HasChildNodes
    {
        get { return this.ChildNodes.Count != 0; }
    }


    /// <summary>
    /// Travearse the Graph recursively
    /// </summary>
    /// <param name="root">root node</param>
    /// <param name="visitor">visitor for each node</param>
    public void Traverse(Node<T> root, Action<Node<T>> visitor)
    {
        if (root == null)
        {
            throw new ArgumentNullException("root");
        }
        if (visitor == null)
        {
            throw new ArgumentNullException("visitor");
        }

        visitor(root); 
        foreach (var node in root.ChildNodes)
        {
            Traverse(node, visitor);
        }
    }

    /// <summary>
    /// Value of the node
    /// </summary>
    public T Value { get; private set; }

    /// <summary>
    /// List of all child nodes
    /// </summary>
    public List<Node<T>> ChildNodes { get; private set; }
}

这很简单。方法:

/// <summary>
/// Helper class for Node 
/// </summary>
/// <typeparam name="T">Value of a node</typeparam>
public static class NodeHelper
{
    /// <summary>
    /// Converts Directed Acyclic Graph to Tree data structure using recursion.
    /// </summary>
    /// <param name="root">root of DAG</param>
    /// <param name="seenNodes">keep track of child elements to find multiple connections (f.e. A connects with B and C and B also connects with C)</param>
    /// <returns>root node of the tree</returns>
    public static Node<T> DAG2TreeRec<T>(this Node<T> root, HashSet<Node<T>> seenNodes)
    {
        if (root == null)
        {
            throw new ArgumentNullException("root");
        }
        if (seenNodes == null)
        {
            throw new ArgumentNullException("seenNodes");
        }

        var length = root.ChildNodes.Count;
        for (int i = 0; i < length; ++i)
        {
            var node = root.ChildNodes[i];
            if (seenNodes.Contains(node))
            {
                var nodeClone = new Node<T>(node.Value, node.ChildNodes);
                node = nodeClone;
            }
            else
            {
                seenNodes.Add(node);
            }
            DAG2TreeRec(node, seenNodes);
        }
        return root;
    }
    /// <summary>
    /// Converts Directed Acyclic Graph to Tree data structure using explicite stack.
    /// </summary>
    /// <param name="root">root of DAG</param>
    /// <param name="seenNodes">keep track of child elements to find multiple connections (f.e. A connects with B and C and B also connects with C)</param>
    /// <returns>root node of the tree</returns>
    public static Node<T> DAG2Tree<T>(this Node<T> root, HashSet<Node<T>> seenNodes)
    {
        if (root == null)
        {
            throw new ArgumentNullException("root");
        }
        if (seenNodes == null)
        {
            throw new ArgumentNullException("seenNodes");
        }

        var stack = new Stack<Node<T>>();
        stack.Push(root);

        while (stack.Count > 0) 
        {
            var tempNode = stack.Pop();
            var length = tempNode.ChildNodes.Count;
            for (int i = 0; i < length; ++i)
            {
                var node = tempNode.ChildNodes[i];
                if (seenNodes.Contains(node))
                {
                    var nodeClone = new Node<T>(node.Value, node.ChildNodes);
                    node = nodeClone;
                }
                else
                {
                    seenNodes.Add(node);
                }
               stack.Push(node);
            }
        } 
        return root;
    }
}

和测试:

    static void Main(string[] args)
    {
        // Jitter preheat
        Dag2TreeTest();
        Dag2TreeRecTest();

        Console.WriteLine("Running time ");
        Dag2TreeTest();
        Dag2TreeRecTest();

        Console.ReadKey();
    }

    public static void Dag2TreeTest()
    {
        HashSet<Node<int>> hashSet = new HashSet<Node<int>>();

        Node<int> root = BulidDummyDAG();

        Stopwatch stopwatch = new Stopwatch();
        stopwatch.Start();
        var treeNode = root.DAG2Tree<int>(hashSet);
        stopwatch.Stop();

        Console.WriteLine(string.Format("Dag 2 Tree = {0}ms",stopwatch.ElapsedMilliseconds));

    }

    private static Node<int> BulidDummyDAG()
    {
        Node<int> node2 = new Node<int>(2);
        Node<int> node4 = new Node<int>(4);
        Node<int> node3 = new Node<int>(3);
        Node<int> node5 = new Node<int>(5);
        Node<int> node6 = new Node<int>(6);
        Node<int> node7 = new Node<int>(7);
        Node<int> node8 = new Node<int>(8);
        Node<int> node9 = new Node<int>(9);
        Node<int> node10 = new Node<int>(10);
        Node<int> root  = new Node<int>(1);

        //making DAG                   
        root.ChildNodes.Add(node2);    
        root.ChildNodes.Add(node3);    
        node3.ChildNodes.Add(node2);   
        node3.ChildNodes.Add(node4);   
        root.ChildNodes.Add(node5);    
        node4.ChildNodes.Add(node6);   
        node4.ChildNodes.Add(node7);
        node5.ChildNodes.Add(node8);
        node2.ChildNodes.Add(node9);
        node9.ChildNodes.Add(node8);
        node9.ChildNodes.Add(node10);

        var length = 10000;
        Node<int> tempRoot = node10; 
        for (int i = 0; i < length; i++)
        {
            var nextChildNode = new Node<int>(11 + i);
            tempRoot.ChildNodes.Add(nextChildNode);
            tempRoot = nextChildNode;
        }

        return root;
    }

    public static void Dag2TreeRecTest()
    {
        HashSet<Node<int>> hashSet = new HashSet<Node<int>>();

        Node<int> root = BulidDummyDAG();

        Stopwatch stopwatch = new Stopwatch();
        stopwatch.Start();
        var treeNode = root.DAG2TreeRec<int>(hashSet);
        stopwatch.Stop();

        Console.WriteLine(string.Format("Dag 2 Tree Rec = {0}ms",stopwatch.ElapsedMilliseconds));
    }

此外，数据结构需要改进:

• 覆盖 GetHash、toString、Equals、== 运算符
• 实现 IComparable
• LinkedList 可能是更好的选择

此外，在转换之前还有一些需要检查的问题:

• 多图
• 如果是 DAG(循环)
• DAG 中的钻石
• DAG 中的多个根

总而言之，它归结为几个问题: 我怎样才能提高转换率？因为这是一个递归，所以有可能炸毁堆栈。我可以添加堆栈来记住它。如果我采用延续传球风格，我会更有效率吗？

我觉得在这种情况下不可变数据结构会更好。是否正确？

Childs 的名字对吗？ :)

最佳答案

算法:

• 如您所见，一些节点在输出中出现了两次。如果节点 2 有 child ，则整个子树将出现两次。如果希望每个节点只出现一次，替换

  if (hashSet.Contains(node))
  {
      var nodeClone = new Node<T>(node.Value, node.Childs);
      node = nodeClone;
  }
  

  与

  if (hashSet.Contains(node))
  {
      // node already seen -> do nothing
  }
  

• 我不会太担心堆栈的大小或递归的性能。但是，您可以将深度优先搜索替换为 Breadth-first-search这将导致更早访问更靠近根的节点，从而产生更“自然”的树(在您的图片中，您已经按 BFS 顺序对节点进行了编号)。

   var seenNodes = new HashSet<Node>();
   var q = new Queue<Node>();
   q.Enqueue(root);
   seenNodes.Add(root);   
  
   while (q.Count > 0) {
       var node = q.Dequeue();
       foreach (var child in node.Childs) {
           if (!seenNodes.Contains(child )) {
               seenNodes.Add(child);
               q.Enqueue(child);
       }
   }
  

  该算法处理菱形和循环。

• 多根

  只需声明一个包含所有顶点的类 Graph

  class Graph
  {
      public List<Node> Nodes { get; private set; }
      public Graph()
      {
          Nodes = new List<Node>();
      }    
  }
  

代码:

• hashSet 可以命名为seenNodes。

• 代替

  var length = root.Childs.Count;
  for (int i = 0; i < length; ++i)
  {
      var node = root.Childs[i];
  

  写

  foreach (var child in root.Childs)
  

• 在 Traverse 中，访问者是完全没有必要的。您可能宁愿有一个方法来生成树的所有节点(以相同的顺序遍历)并且由用户对节点执行任何操作:

  foreach(var node in root.TraverseRecursive())
  {
      Console.WriteLine(node.Value);
  }
  

• 如果您覆盖 GetHashCode 和 Equals，该算法将不再能够区分具有相同值的两个不同节点，这可能不是您想要的。

• 除了 List 在添加节点时所做的重新分配(容量 2、4、8、16，...)之外，我看不出 LinkedList 在这里比 List 更好的任何原因。

关于c# - 将有向无环图 (DAG) 转换为树，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/6397641/