Question

在这段代码中，当我想在两个节点之间添加传递或失败时，我可以轻松地做到这一点。但验证存在问题，如果我们再次创建它创建的相同两个节点之间的边缘。多数民众赞成我不想要的。我只需要私有类ConnectAction在同一个两个节点之间创建一条边。我在那里尝试了一些东西，但它不起作用。如果我们删除迭代器并只创建它被创建的边缘。请帮助。

这是节点类

private static class Node { 
private Point p;
private int r;
private String NAME;
private String nodeid;
private boolean selected = false;
private Rectangle b = new Rectangle();

/**
 * Construct a new node.
 */
public Node(String nodeid,Point p, int r,String NAME) {
    this.nodeid=nodeid;
    this.p = p;
    this.r = r;
    this.NAME = NAME;

    setBoundary(b);

}

/**
 * Calculate this node's rectangular boundary.
 */
private void setBoundary(Rectangle b) {
    b.setBounds(p.x - r, p.y - r, 2 * r, 2 * r);
}

/**
 * Draw this node.
 */
public void draw(Graphics g) {
    g.setColor(Color.white);

            g.fillRect(b.x, b.y, b.width, b.height);
            g.setColor(Color.black);
            g.drawString(NAME,b.x+25,b.y+40);

    if (selected) {
        g.setColor(Color.darkGray);
        g.drawRect(b.x+3, b.y+3, b.width, b.height);
    }
}

/**
 * Return this node's location.
 */
public Point getLocation() {
    return p;
}

/**
 * Return true if this node contains p.
 */
public boolean contains(Point p) {
    return b.contains(p);
}

/**
 * Return true if this node is selected.
 */
public boolean isSelected() {

    return (selected);

}

/**
 * Mark this node as selected.
 */
public void setSelected(boolean selected) {
    this.selected = selected;
}

/**
 * Collected all the selected nodes in list.
 */
public static void getSelected(List<Node> list, List<Node> selected) {
    selected.clear();
    for (Node n : list) {
        if (n.isSelected()) {
            selected.add(n);
        }
    }
}

/**
 * Select no nodes.
 */
public static void selectNone(List<Node> list) {
    for (Node n : list) {
        n.setSelected(false);
    }
}

/**
 * Select a single node; return true if not already selected.
 */
public static boolean selectOne(List<Node> list, Point p) {
    for (Node n : list) {
        if (n.contains(p)) {
            if (!n.isSelected()) {
                Node.selectNone(list);
                n.setSelected(true);
            }
            return true;
        }
    }
    return false;
}

/**
 * Select each node in r.
 */
public static void selectRect(List<Node> list, Rectangle r) {
    for (Node n : list) {
        n.setSelected(r.contains(n.p));

    }
}

/**
 * Toggle selected state of each node containing p.
 */
public static void selectToggle(List<Node> list, Point p) {
    for (Node n : list) {
        if (n.contains(p)) {
            n.setSelected(!n.isSelected());
        }
    }
}

/**
 * Update each node's position by d (delta).
 */
public static void updatePosition(List<Node> list, Point d) {
    for (Node n : list) {
        if (n.isSelected()) {
            n.p.x += d.x;
            n.p.y += d.y;
            n.setBoundary(n.b);

        }
    }
}

}

这是边缘类

    private static class Edge{

private Node n1;
private Node n2;
private String Ctype;
private String eid;

private static String edgeid;

public Edge(String eid,Node n1, Node n2, String Ctype) {
    this.eid = eid;
    this.n1 = n1;
    this.n2 = n2;
    this.Ctype=Ctype;


}

public void draw(Graphics g) {
    Point p1 = n1.getLocation();
    Point p2 = n2.getLocation();
    g.setColor(Color.darkGray);
    g.drawLine(p1.x, p1.y, p2.x, p2.y);
    g.drawString(Ctype,(p1.x+p2.x)/2,(p1.y+p2.y)/2);
}

}

这里面临的问题是在同一对节点之间只创建一条边。

private class ConnectAction extends AbstractAction {

String ConnectionType;

public ConnectAction(String name) {
    super(name);
    this.ConnectionType=name;
}

public void actionPerformed(ActionEvent e) {
    Node.getSelected(nodes, selected);
    if (selected.size() > 1) {
        for (int i = 0; i < 1; ++i) {
            Node n3 = selected.get(i);
            Node n4 = selected.get(i + 1);
           String a= n3.nodeid;
           String b= n4.nodeid;
          // System.out.println(a);
           //System.out.println(b);

           //System.out.println(Edgeid);


           ListIterator<Edge> iter = edges.listIterator();
           while (iter.hasNext()) {
               Edge e1 = iter.next();
               if (!(n3.nodeid == e1.n1.nodeid && n4.nodeid == e1.n2.nodeid) || (n4.nodeid == e1.n1.nodeid && n3.nodeid == e1.n2.nodeid)) {
                   String Edgeid=ID.giveid();
                   Edge ed=new Edge(Edgeid,n3, n4,ConnectionType);
                   edges.add(ed);

               }iter.remove();

           }                                                                                                                                                                                                                                                                         }                                                             
    }     
    repaint();
}

Answer 1

也许我误解了你想要的东西（或你的代码）但是这个

ListIterator<Edge> iter = edges.listIterator();
while (iter.hasNext()) {
    Edge e1 = iter.next();
    if (!(n3.nodeid == e1.n1.nodeid && n4.nodeid == e1.n2.nodeid) || (n4.nodeid == e1.n1.nodeid && n3.nodeid == e1.n2.nodeid)) {
        String Edgeid=ID.giveid();
        Edge ed=new Edge(Edgeid,n3, n4,ConnectionType);
        edges.add(ed);
    }
    iter.remove();
}

创建并删除大量边。（并且您在条件中有错误（！仅适用于第一部分）

我想你应该有这样的事情：

boolean exists = false;
ListIterator<Edge> iter = edges.listIterator();

while (iter.hasNext()) {
    Edge e1 = iter.next();
    if ( (n3.nodeid.equals(e1.n1.nodeid) && n4.nodeid.equals(e1.n2.nodeid)) || (n4.nodeid.equals(e1.n1.nodeid) && n3.nodeid.equals(e1.n2.nodeid)) ) {
        exists = true;

    }
} 

if ( !exists )  {
    String Edgeid=ID.giveid();
    Edge ed=new Edge(Edgeid,n3, n4,ConnectionType);
    edges.add(ed);
}

此外，您通常应使用String.equals（String）来比较字符串。

图的两个节点之间需要单边

1 个答案: