QuickGraph - is there algorithm for find all parents (up to root vertex's) of a set of vertex's

In QuickGraph - is there algorithm for find all parents (up to root vertex's) of a set of vertex's. In other words all vertex's which have somewhere under them (开发者_StackOverflow社区on the way to the leaf nodes) one or more of the vertexs input. So if the vertexs were Nodes, and the edges were a depends on relationship, find all nodes that would be impacted by a given set of nodes.

If not how hard is it to write one's own algorithms?

Here's what I've used to accomplish a predecessor search on a given vertex:

IBidirectionalGraph<int, IEdge<int>> CreateGraph(int vertexCount)
{
    BidirectionalGraph<int, IEdge<int>> graph = new BidirectionalGraph<int, IEdge<int>>(true);
    for (int i = 0; i < vertexCount; i++)
        graph.AddVertex(i);

    for (int i = 1; i < vertexCount; i++)
        graph.AddEdge(new Edge<int>(i - 1, i));

    return graph;
}

static public void Main()
{
    IBidirectionalGraph<int, IEdge<int>> graph = CreateGraph(5);

    var dfs = new DepthFirstSearchAlgorithm<int, IEdge<int>>(graph);            
    var observer = new VertexPredecessorRecorderObserver<int, IEdge<int>>();

    using (observer.Attach(dfs)) // attach, detach to dfs events
        dfs.Compute();

    int vertexToFind = 3;
    IEnumerable<IEdge<int>> edges;
    if (observer.TryGetPath(vertexToFind, out edges))
    {
        Console.WriteLine("To get to vertex '" + vertexToFind + "', take the following edges:");
        foreach (IEdge<int> edge in edges)
            Console.WriteLine(edge.Source + " -> " + edge.Target);
    }
}

Note that if you know your root beforehand, you can specify it in the dfs.Compute() method (i.e. dfs.Compute(0)).

-Doug

I used Doug's answer and found out that if there are more than one parent for a vertex, his solution only provides one of the parents. I am not sure why.

So, I created my own version which is as follows:

    public IEnumerable<T> GetParents(T vertexToFind)
    {
        IEnumerable<T> parents = null;

        if (this.graph.Edges != null)
        {
            parents = this.graph
                .Edges
                .Where(x => x.Target.Equals(vertexToFind))
                .Select(x => x.Source);
        }

        return parents;
    }

You either need to maintain a reversed graph, or create a wrapper over the graph that reverses every edge. QuickGraph has the ReversedBidirectionalGraph class that is a wrapper intended just for that, but it does not seem to work with the algorithm classes because of generic type incompatibility. I had to create my own wrapper class:

class ReversedBidirectionalGraphWrapper<TVertex, TEdge> : IVertexListGraph<TVertex, TEdge> where TEdge : IEdge<TVertex> 
{
  private BidirectionalGraph<TVertex, TEdge> _originalGraph;
  public IEnumerable<TEdge> OutEdges(TVertex v)
    {
        foreach (var e in _graph.InEdges(v))
        {
            yield return (TEdge)Convert.ChangeType(new Edge<TVertex>(e.Target, e.Source), typeof(TEdge));
        }
    } //...implement the rest of the interface members using the same trick
}

Then run DFS or BFS on this wrapper:

var w = new ReversedBidirectionalGraphWrapper<int, Edge<int>>(graph);    
var result = new List<int>();    
var alg = new DepthFirstSearchAlgorithm<int, Edge<int>>(w);
alg.TreeEdge += e => result.Add(e.Target);    
alg.Compute(node);

Doug's answer is not correct, because DFS will only visit the downstream subgraph. The predecessor observer does not help.