Question: Given a weighted directed graph, how can the Dijkstra algorithm be modified to test for the presence of multiple lowest-cost paths between a given pair of nodes?

My current algorithm is as follows: (credit to Weiss)

```
/**
 * Single-source weighted shortest-path algorithm. (Dijkstra) 
 * using priority queues based on the binary heap
 */
public void dijkstra( String startName )
{
    PriorityQueue<Path> pq = new PriorityQueue<Path>( );

    Vertex start = vertexMap.get( startName );
    if( start == null )
        throw new NoSuchElementException( "Start vertex not found" );

    clearAll( );
    pq.add( new Path( start, 0 ) ); start.dist = 0;

    int nodesSeen = 0;
    while( !pq.isEmpty( ) && nodesSeen < vertexMap.size( ) )
    {
        Path vrec = pq.remove( );
        Vertex v = vrec.dest;
        if( v.scratch != 0 )  // already processed v
            continue;

        v.scratch = 1;
        nodesSeen++;

        for( Edge e : v.adj )
        {
            Vertex w = e.dest;
            double cvw = e.cost;

            if( cvw < 0 )
                throw new GraphException( "Graph has negative edges" );

            if( w.dist > v.dist + cvw )
            {
                w.dist = v.dist +cvw;
                w.prev = v;
                pq.add( new Path( w, w.dist ) );
            }
        }
    }
}

```

Answer: 
Replace field `prev`, the link to previous vertex with a collection `prevs`, and change the code slightly:

```
...

        if( w.dist >= v.dist + cvw ) {
            if ( w.dist > v.dist + cvw ) {
                w.dist = v.dist +cvw;
                w.prevs.clear();
            }
            w.prevs.add(v);
            pq.add( new Path( w, w.dist ) );
        }

...

```
If you only want to find a single other equal-cost path
-------------------------------------------------------

Let's suppose you already ran Dijkstra's algorithm once to get a shortest path `P`. You can add a tiny cost `epsilon` to each edge in `P` and run Dijkstra's a second time on the modified graph to get a new path `P'`. If `P` and `P'` contain the same edges, then you can conclude that `P` is the unique shortest path. Otherwise, we undo the `epsilon` change and compare the lengths of `P` and `P'`. If the lengths are equal, then clearly `P'` is another distinct shortest path. Otherwise, `P` is the unique shortest path.

If we want to find all shortest paths
-------------------------------------

Such an algorithm would necessarily be exponential-time. This is because a graph can have exponentially-many equal-cost paths between two nodes. For example, consider the graph:

```
A --> B1 --> C --> D1 --> E ...
  \       ->   \       ->
   -> B2 /      -> D2 /

```

There are 4 paths from `A` to `E`, and if we assume that all edges are equal cost, then all of those paths have equal total cost. By repeating this pattern we can get exponentially-many paths of equal cost.