I found rng58's analysis, even muted, pretty easy to understand: https://youtu.be/WFg2yJGZ2Cw?t=47m41s

My solution:

• we can find N subtournaments of 2⁰ to 2^N - 1 players; player 1 plays against the winner — the smallest player — from each group, we need to ensure neither of them is one of the given M

• there are too many choices of these smallest opponents, but we can use inclusion-exclusion — we only need to compute the number of tournaments in which the winners of |S| subtournaments belong to a subset S (of the given M opponents) for each |S|

• if we fixed S, fixed the sizes of subtournaments won by members of S and went in decreasing order of numbers in S, this number can be computed as a combinatorial formula: for a given , if the subtournaments for previous (larger) numbers in S have total size x and its subtournament has size 2^k, we're choosing 2^k - 1 numbers out of 2^N - a - x (all larger unused numbers) and then we have (2^k)! ways to order them within the subtournament; at the end, there are 2^N - 1 - x unused remaining numbers and we can choose their order in (2^N - 1 - x)! ways

• note that we didn't use the actual values larger than a in S at all, only the sum of sizes of their subtournaments, so we can turn it into a DP that counts the number of ways to use j numbers out of A_i..M as smallest numbers in subtournaments with total size x

• since the subtournaments' sizes are powers of 2, x is also the bitmask of already used subtournaments, so we can transition from a state (i + 1, j, x) to (i, j + 1, x + 2^k) if the k-th bit in x is unset

• this gives O(M²2^N) states and O(M²N2^N) transitions, each of which can be handled in O(1) if we precompute factorials and their inverses

Presumably a similar idea: https://pastebin.com/XYvXUtXR

I will try to explain my idea.
I calculated for every pair of edges the min spanning tree cost containing those two edges.This can be done by using n min spanning trees for each edge and constructing the tree and using max queries on paths of trees.
Let val[i][j] denote min spanning tree cost containing ith and jth edge.now if val[i][j] < x then they both should have same colour.so join then (I used dsu here). After doing this we will get some components in which each must be colored with a single color.so 2^components will give number of ways to get min spanning tree more than equal to x.now similarly do last step assuming x=x+1.and subtract both which is required answer.

It would be good if someone related posts a blog post here before contest.Last week I thought it was on Saturday(I was busy on Sat) and I missed it. But one of my friends said it is Sunday.