I am implementing this algorithm for a directed graph. But the interesting thing about this graph nodes have also their own flow capacities. I think, this subtle change of the o
There's a simple reduction from the max-flow problem with node capacities to a regular max-flow problem:
For every vertex v in your graph, replace with two vertices v_in and v_out. Every incoming edge to v should point to v_in and every outgoing edge from v should point from v_out. Then create one additional edge from v_in to v_out with capacity c_v, the capacity of vertex v.
So you just run Edmunds-Karp on the transformed graph.
So let's say you have the following graph in your problem (vertex v has capacity 2):
s --> v --> t
1 2 1
This would correspond to this graph in the max-flow problem:
s --> v_in --> v_out --> t
1 2 1
It should be apparent that the max-flow obtained is the solution (and it's not particularly difficult to prove either).