You are right about that. I think repeatedly removing leaves of the graph until only nodes of degree $$$\geq 2$$$ are left will reduce the amount of paths by an order of $$$n$$$, since the sum of degrees of all nodes will be $$$\leq 2n+20$$$, and the maximum degree of any node will also be 20. It should pass within the time limit now.

390263643

Alternative solution for F:

After seeing that the edges colored 1 and 2 must form a cycle, we also observe that this cycle must have length $$$\leq m-n+1$$$. This is because the graph still needs to be connected after removing the cycle. The minimum number of edges needed for a graph of size $$$n$$$ to be connected is $$$n-1$$$ edges, so if the cycle was more than $$$m-n+1$$$ edges long, it is impossible for the graph to be connected after removing them.

So, we just run a dfs starting from each node, with a max depth of 10, finding every cycle candidate containing that node. The complexity should be the same as in the official solution.

Submission: 390181201