### Блог пользователя Maxon5

Автор Maxon5, история, 11 месяцев назад, ,

I've been trying to understand Dinic's Max Flow algorithm. I think I've fairly understood how the algorithm works.

In each iteration, you create a level graph using BFS and then find blocking flow in that level graph. You would need O(N) iterations and in each iteration, blocking flow can be found in O(NM) complexity. This is the place where I'm stuck. I've seen some implementations like Stanford Notebook Dinic implementation . But it seems to me that in the worst case you might need O(M^2) complexity to find the blocking flow if you implement the code this way. Can anyone help me to understand why the complexity is O(NM) for finding blocking flow ?

•
• +25
•

 » 11 месяцев назад, # |   +10 Hello random people downvoting a valid question. Care to explain ?
 » 11 месяцев назад, # |   +20 First you run bfs which is O(M)Then you run dfs which would run in O(M) but sometimes you return to the origin (when you find path to the end). Suppose between to returns to origin you visited X edges and found path from s to t is Y. It means that you spent X time, and will never run through other X - Y edges (Note that to really skip them there's array of pt in the code)So, overall time will be = But is O(NM) because each Y does not exceed N and you do at most O(M) such sub-iterations, because each time you do it, you fill an edge completely.
•  » » 11 месяцев назад, # ^ |   0 Thanks a lot. This makes sense.I was actually missing the part of array of pt in the code.
•  » » 11 месяцев назад, # ^ |   0 Another thing, I can intuitively understand that in each successive iteration, the length of the shortest path from source to sink would increase. So, since the length can never be more than V, we would need O(V) iterations. Can you please give me some idea about how to prove concretely that the length always increases in each successive iteration?
•  » » » 11 месяцев назад, # ^ | ← Rev. 2 →   +15 1) Prove that distance will not decrease for any vertex2) Prove what you needMore elaborate plan in first edit
•  » » » » 11 месяцев назад, # ^ |   0 Thanks again. I've got the idea.
 » 11 месяцев назад, # |   +13 A little bit off topic. I found this code in which it is said that it runs in O(N * M * log(MC)), where MC is maximum edge capacity. Can someone explain why such a complexity?
•  » » 11 месяцев назад, # ^ |   +5 This is interesting. I've heard about using dynamic trees to get the above mentioned complexity but this code seems fairly simple. Can you please explain the idea behind it, adja ?
•  » » » 11 месяцев назад, # ^ |   +5 It is standart scaling technique — limiting flow size from up to down. It also can be used in Ford-Fulkerson algorithm.
•  » » 11 месяцев назад, # ^ | ← Rev. 3 →   0 I've checked both ADilet JAksybay's implementation and my own O(N^2 * M) implementation for the problem SPOJ FASTFLOW. The runtime of my code is 0.09 seconds whereas the same code with ADilet JAksybay's implementation of Dinic is taking 1.28 seconds. I'm now a bit confused if the complexity stated in the comments of ADilet JAksybay's code is actually correct. This is submission with my implementation.This is submission with ADilet JAksybay's implementation.
•  » » » 11 месяцев назад, # ^ |   0 Actually his name is ADilet JAksybay