### game_changer007's blog

By game_changer007, history, 3 years ago, ,

Hi,I am trying to solve this problem on Spoj.

I have implemented simple Dinic's algorithm only,but am getting TLE.Here is the solution link solution .Am I doing anything wrong?

Edit:Got AC!!

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 » 3 years ago, # |   +6 In your dfs function you are unnecessarily iterating over the used edges. You only need to start from scratch whenever you recompute the bfs function(), and not everytime.In short1)Make a ptr[] array (size >= number of vertices)2)change line 75 to for(;ptr[u]
•  » » 3 years ago, # ^ |   +5 AC :) Thanks... Learned something new!!
•  » » » 3 years ago, # ^ |   0 Welcome :)
•  » » » » 3 years ago, # ^ |   0 Hi!! Another question..Is Dinic's algorithm the same as Hopcroft-Karp algorithm?? Do we need to make any changes in Dinic's to make it Hopcroft-Karp?If yes,then how and in which part?
•  » » » » » 3 years ago, # ^ | ← Rev. 2 →   -8 I wasn't right.
•  » » » » » 3 years ago, # ^ |   +9 Actually, Hopcroft-Karp is an algorithm to find the maximum matching in a bipartite graph (and it has the same time complexity as Dinic's algorithm on bipartite graphs, but it's much faster in practice). You can always implement bipartite matching with Dinic's, but Hopcroft-Karp is easier to code in my opinion.
•  » » » » » » 3 years ago, # ^ | ← Rev. 2 →   +1 Hi!! As you said that Dinic's has the same complexity as Hopcroft-Karp for maximum matching... But am getting TLE for this problem on applying Dinic's algorithm.My solution is this.Am i missing anything??? I read that HK is accepted in this problem..So should i conclude that HK is faster than Dinic's in general?
•  » » » » » » » 3 years ago, # ^ |   0 Dinic's has the same complexity as Hopcroft-Karp in bipartite graphs. I guess you should do constant optimizations (because there are people who got AC with Dinic's) or change stuff such as the order of some fors (n -> 1 instead of 1 -> n). If all else fails, implement Push-Relabel (Which is a guaranteed AC with heuristics, but much painful to code).
•  » » 23 months ago, # ^ |   0 can you explain the statement fl=mini(fl,cap[u][to]-flow[u][to]); in the code.You are passing the bottleneck of one of the edge into all the children
 » 3 years ago, # |   0 Another way of changing it is just adding something like d[u] = 1000000000ll before return 0` in the dfs.Here is an accepted solution of your code: http://ideone.com/LBl7V9
•  » » 3 months ago, # ^ |   0 why did that work ?? i cant understand