Maximum number of matched vertexes in a one-to-many bipartite graph

I have a variant of bidding problem at hand.
There are N bidders(~20) who bid for items from a pool of many items(~10K). Each bidder can bid many items. I want to maximize the number of bidders who are satisfied. A bidder is satisfied if he gets all the items that he bid for in the first place. For eg-

Bidders = A,B,C
Items = 1,2,3,4

Bidder    Bids
A         1,2
B         2,3
C         3,4

In this case its only possible to satisfy 2 bidders at max.

I've tried to model the problem to a maxflow problem and have taken several approaches but to no avail
My approaches so far-

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Tried to model this problem as a bipartite matching problem. The only problem being that instead of a one-one mapping I have a one-many mapping with an AND condition.

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A maxflow problem with edges going from source to each vertex with a capacity of number of bids. Problem here being ensuring that all edges from a bidder are selcted.

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A maxflow problem with both upper bounded and lower bounded edge capacities.

Since your problem is exactly the set packing problem, it is NP-hard. However, you only have $n = 20$ bidders, so you can solve it using a simple exponential algorithm:

• For every item $i$, build the subset $B_i$ of bidders that bid for it. You can represent this subset for example as a bitmask. You know that for all $i$, every subset $S \subset B_i$ with $|S| > 1$ can not be a subset of the solution.

• For all $B_i$, enumerate all of its "forbidden" subsets of size > 1 and store them. Avoid revisiting already stored subsets by using recursion and memoization, by canceling the bits one after another.

• Go through all subsets of bidders. For every set, check whether it has a a forbidden subset. You can use the same recursive algorithm as in step 2.

Note that if you only enumerate subsets of size 2 in step 2, you have reduced your problem to an independent set problem in a graph with $n$ nodes, so you can apply any other more efficient algorithm to it.

Runtime: $O(n \cdot (2^n + m))$, with $m$ being the number of items and $n$ being the number of bidders. With a good independent set algorithm you can get something like $O(m \cdot n^2 + 1.3^n)$

StackExchange Computer Science Q#22542, answer score: 2