在分支中達到頂點的第n個后繼

Question

我使用jgrapht庫創建了一個有向圖。 我使用successorListOf()方法訪問頂點的后繼者，但我希望能夠到達給定頂點的第n個后繼者（在我的例子中是Point對象）。 我的有向圖有兩個分支（這里分別命名為B和C）。 我編寫了一個簡單而簡短的代碼以使其變得更容易：

public static DirectedGraph<Point, DefaultEdge> directedGraph = new DefaultDirectedGraph<Point, DefaultEdge>(DefaultEdge.class);
public static Point startPoint = new Point(2, 6, "S");
public static Point firstPoint = new Point(2, 7, "A");
public static Point secondPoint = new Point(2, 8, "B");
public static Point thirdPoint = new Point(2, 9, "B");
public static Point fourthPoint = new Point(2, 10, "B");
public static Point fifthPoint = new Point(3, 7, "C");
public static Point sixthPoint = new Point(4, 7, "C");
public static Point seventhPoint = new Point(5, 7, "C");


void setup ()  {
  directedGraph.addVertex(startPoint);
  directedGraph.addVertex(firstPoint);
  directedGraph.addVertex(secondPoint);
  directedGraph.addVertex(thirdPoint);
  directedGraph.addVertex(fourthPoint);
  directedGraph.addVertex(fifthPoint);
  directedGraph.addVertex(sixthPoint);
  directedGraph.addVertex(seventhPoint);
  directedGraph.addEdge(startPoint, firstPoint);
  directedGraph.addEdge(firstPoint, secondPoint);
  directedGraph.addEdge(firstPoint, thirdPoint);
  directedGraph.addEdge(firstPoint, fourthPoint);
}

// --------------------------------------------------------------
public static ArrayList<Point> pointList = new ArrayList<Point>();
public static class Point {

  public int x;
  public int y;
  public String iD;
  public  Point(int x, int y, String iD) 
  {

    this.x = x;
    this.y = y;
    this.iD= iD;
  }
  @Override
    public String toString() {
    return ("[x="+x+" y="+y+" iD="+iD+ "]");
  }

  @Override
    public int hashCode() {
    int hash = 7;
    hash = 71 * hash + this.x;
    hash = 71 * hash + this.y;

    return hash;
  }



  @Override
    public boolean equals(Object other) 
  {
    if (this == other)
      return true;

    if (!(other instanceof Point))
      return false;

    Point otherPoint = (Point) other;
    return otherPoint.x == x && otherPoint.y == y;
  }
}

我想在firstPoint和“ B”分支的每個點之間添加一條邊，而不是：

directedGraph.addEdge(firstPoint, secondPoint);
  directedGraph.addEdge(firstPoint, thirdPoint);
  directedGraph.addEdge(firstPoint, fourthPoint);

我想使用：

for (Point successor : Graphs.successorListOf (directedGraph, firstPoint)) {
        if (successor.type.equals("B") {
               directedGraph.addEdge(firstPoint, successor);
        }
}

但是在這里，我只能到達分支B的第一個后繼者。對於第n個后繼者，我如何到達后繼者等的后繼者？ B分支中的頂點數量可能會發生變化，這就是為什么我正在尋找一種自動執行此操作的方法，而不是逐點進行操作的原因。

我該怎么辦？

在圖形上，1將是我的起點，2將是我的firstPoint，然后有兩個分支將是我的B＆C分支

Answer 1

我編寫了以下代碼，但未經測試，您可能需要對其進行修改以滿足您的要求。

此代碼使用您在提供的示例中使用的變量和實例將DFS（深度優先搜索）運行到預定義的深度。

public void getSuccessor(DirectedGraph<Point, DefaultEdge> graph, Point point, String type, int depth) {
    List<Point> visitedPoints = new ArrayList<>();
    _getSuccessor(graph, point, visitedPoints, type, depth);
}

private void _getSuccessor(DirectedGraph<Point, DefaultEdge> graph, Point point, List<Point> visitedPoints, String type, int depth){

    if(depth == 0)
        return;

    // Mark node as visited
    visitedPoints.add(point);

    // Loop on all its successors
    for(Point successor : Graphs.successorListOf (directedGraph, point)){

        // If node not already visited
        if(!visitedPoints.contains(successor) && successor.type.equals(type)) {
            directedGraph.addEdge(firstPoint, successor);
            _getSuccessor(graph, successor, visitedPoints, type, depth-1);
        }
    }
}