Hello! Can somebody tell me if there is a faster algo than O(Q^2*N)? My algo works pretty fast, but I wondered if there is a better way to do it.

My solution Q*N, lol
Or no
I'm not sure

Edited by author 11.03.2016 21:50

Edited by author 11.03.2016 21:50

My algo is fills dp[vertex][presaved edges] array (result is dp[1][Q]), where total loops N^3/6 <  2*10^5.
There maximum presaved edges = Q, but more vertexes have less edges than Q.
Let edges[v] - number of edges of v vertex.

So there  O( N *  sum( edges[ v ] * ( edges[ v ] + 1 ) / 2 ) ) = O( N^3 / 6 ).