Floyd-Warshall: all shortest paths

Question

I\'ve implemented Floyd-Warshall to return the distance of the shortest path between every pair of nodes/vertices and a single shortest path between each of the

Answer 1

The 'counting' function in the current approved answer flails in some cases. A more complete solution would be:

procedure FloydWarshallWithCount ()
for k := 1 to n
    for i := 1 to n
        for j := 1 to n
            if path[i][j] == path[i][k]+path[k][j]
                count[i][j] += count[i][k] * count[k][j]
            else if path[i][j] > path[i][k] + path[k][j]
                path[i][j] = path[i][k] + path[k][j]
                count[i][j] = count[i][k] * count[k][j]

The reason for this is that for any three vertices i, j, and k, there may be multiple shortest paths that run from i through k to j. For instance in the graph:

       3             1
(i) -------> (k) ---------> (j)
 |            ^
 |            |
 | 1          | 1
 |     1      |
(a) -------> (b)

Where there are two paths from i to j through k. count[i][k] * count[k][j] finds the number of paths from i to k, and the number of paths from k to j, and multiplies them to find the number of paths i -> k -> j.