Algorithms Graph Definition – Minimum Cost Graph in a Strongly Connected Component

Yes, both of your premises describe necessary conditions for a cyclic graph. However, they only exclude trivial graphs from cycle analysis, and you'll want to use one of the standard algorithms to make sure a given graph is actually cycle-free.

There are actually fairly cheap algorithms for detecting cycles in graphs; if you are performing a topological sort on a graph, for example, you'll detect cycles during the sort. The exact choice of algorithm naturally depends on your graph (one outgoing edge or multiple?) and your requirements.

Here's an implementation that I dug up from when I was dealing with graphs. Although it's in C#, with a few minor adjustments it could compile in Java.

To make it directed you would have to copy v.Adjacents.Add(new Edge(w, cost)); and reverse the direction, thereby taking up double the space.

I don't think it's any better than what you have though.

class Edge
{
    public Vertex Destination { get; set; }
    public double Cost { get; set; }

    public Edge(Vertex destination, double cost)
    {
        Destination = destination;
        Cost = cost;
    }

}

class Vertex
{
    public string Name { get; set; }
    public List<Edge> Adjacents { get; set; }
    public double Distance { get; set; }

    public Vertex(string name)
    {
        Name = name;
        Adjacents = new List<Edge>();
        Distance = double.MaxValue;
    }
}

class Graf
{
    private readonly Dictionary<string, Vertex> vertexmap = new Dictionary<string, Vertex>();

    public void AddEdge(string source, string dest, double cost)
    {
        var v = GetVertex(source);
        var w = GetVertex(dest);
        v.Adjacents.Add(new Edge(w, cost));
    }

    private Vertex GetVertex(string vertexname)
    {
        Vertex v;
        vertexmap.TryGetValue(vertexname, out v);
        if (v == null)
        {
            v = new Vertex(vertexname);
            vertexmap.Add(vertexname, v);
        }
        return v;
    }
}