如何知道我的图中是否存在顶点？

Question

I'm making a method that load a graph from a file. It's very simple, but if there are repeated vertex, the method'll insert too into the graph, so I'm trying to avoid this.

This is my current code:

public static Graph<ElementoDecorado<Integer>, String> loadGraphFromFile(File f) {
        boolean v1_exists = false, v2_exists = false;
        Graph<ElementoDecorado<Integer>, String> g = new AdjacencyListGraph<ElementoDecorado<Integer>, String>();
        Vertex<ElementoDecorado<Integer>> v1, v2, aux = null;
        Scanner fr;

        try {
            fr = new Scanner(f);

            while(fr.hasNextLine()) {
                v1 = g.insertVertex(new ElementoDecorado<Integer>(fr.nextInt()));
                v2 = g.insertVertex(new ElementoDecorado<Integer>(fr.nextInt()));

                for(Vertex<ElementoDecorado<Integer>> v : g.vertices()) {
                    if(v.equals(v1)) {

                        /*aux = v;
                        v1_exists = true;*/
                    }
                    if(v.equals(v2)) {
                        /*aux = v;
                        v2_exists = true;*/
                    }
                }

                g.insertEdge(v1, v2, "edge");

                v1_exists = v2_exists = false;
            }
        } catch(FileNotFoundException e) {
            e.printStackTrace();
        }

        return g;
    }

I don't know what to write into the two ifs. I have tried to delete the vertex if they are equal but obviously this doesn't work cause at the end my graph'll be empty :S

This is the manual page for the Vertex interface.

Any help is welcome. Thanks, and merry christmas!

Answer 1

You should first check what graph.insertVertex(V value) does. If the package was build decently (which I doubt from the poor documentation), then that method will only create a new vertex if a vertex with value does not already exist; otherwise it returns the existing vertex of value value .

However I can't tell from the non-documentation whether the package really assumes that there is a single vertex for a given value and whether insertVertex behaves correctly.

Here is some code in case insertVertex does not check for duplication:

(I replaced ElementoDecorado<Integer> by Integer for readability)

while(fr.hasNextLine()) {
   int nodeId1 = fr.nextInt();
   int nodeId2 = fr.nextInt();
   Vertex<Integer> vert1 = null;
   Vertex<Integer> vert2 = null;

   for(Vertex<Integer> v : g.vertices()) {
      int nodeId = v.element();
      if(nodeId == nodeId1) 
         vert1 = v;
      else if (nodeId == nodeId2) // assumes can't have nodeId1 == nodeId2
         vert2 = v;
    }
    if (vert1 == null)
       vert1 = g.insertVertex(nodeId1);
    if (vert2 == null)
       vert2 = g.insertVertex(nodeId2);

    g.insertEdge(vert1, vert2, null);
 }

Answer 2

Before you insert the vertex in the graph, ask if there is a already a String value for the given vertex key.

vertex = new ElementoDecorado<Integer>(fr.nextInt());
if(graph.get(key) != null) { 
    //it exists, don't insert it
} else { 
    g.insertVertex(vertex)
}

Know your own data structures. Ask yourself, what is the graph made of? It's just a mapping of :

ElementoDecorado<Integer> => String

Also do not name your variables things like: g . It doesn't convey any meaning.

Answer 3

The problem is that you're adding the vertices to the graph BEFORE you check if you need to add them. First, instead of directly adding both vertices, just read them:

v1 = new ElementoDecorado<Integer>( fr.nextInt() );
v2 = new ElementoDecorado<Integer>( fr.nextInt() );

Then, in the for , you check whether the vertex exists, just like you were trying. Your if s could look like this:

if( !v1_exists && v.equals( v1 ) ) {
    v1 = v;
    v1_exists = true;
}

And similarly for v2 . At the end, if and only if v1_exists is false, you add the vertex:

// right anfter the for-each
if ( !v1_exists ) {
    g.addVertex( v1 );
}
if ( !v1_exists ) {
    g.addVertex( v2 );
}

g.insertEdge( v1, v2, "edge" );
// etc

There are a few optimizations you could also do, such as stopping the for when both vertices are found, but that should do it.

Note that it might be a better way of doing it, but I don't know those classes.