Thanks for translation by RodionGork

Problem statement:
Given an oriented graph with N vertices and M edges we are to find all vertex pairs for which exists a path of infinitesimal weight. 

About 2 months ago I wrote an article in russian version on algorithm for this problem. After some more work and with the help of new idea provided by Narg, I created an improved version of the algorithm. Here is its description 

Step 1. Build condensation of the graph. For each of strongly connected components (SCC) check, whether it contains any loop of negative weight - if so, paint all vertices of this SCC black - otherwise paint them white. 

Now note that if any strongly connected component (SCC) contains a negative-weight loop, then any pair of vertices from this component have infinitesimal-length path between them, since when building a path for any pair we could enter negative-weight loop and reduce total weight to any value. 

Use Ford-Bellman algorithm for each of SCC to find out whether it contains negative-weight loop. 

Step 2. For each of vertices perform BFS from it. If some vertex B could be reached from given starting vertex A via path, containing any black intermediate vertex - then the vertex pair (A, B) is one of those we seek for. Remember it or print out. Working this way we could find out all such pairs. 

Time complexity, by steps, is O((N1 + N2 + ... + Nk)*M) = O(N*M) for condensation and Ford-Bellman and O(N*M) for BFS. Summary does not exceed O(N*M).

Memory: O(n + m).

There is the code

I've tested my algorithm against implementation of Floyd by Edward Davtyan. I performed testing for three kinds of graphs: containing almost full set of edges, containing about quarter of this amount of edges and containing only about 1/20 of full set of edges. 

There is the test generator

Results: