Algorithm for picking items from sets

Question

I have a collection of sets S[i] , I need to pick C[i] items from each correspoinding set. Some items might belong to several sets at once, picking the same item twice is not allowed.

Here's an example to explain better:

Set #1 [b, c, d, e], pick 2
Set #2 [a, b, c], pick 2
Set #3 [w, x, y, z], pick 1
Set #4 [a, d, e], pick 1

One of the solutions would be:

Set #1 [b, d]
Set #2 [a, c]
Set #3 [x]
Set #4 [e]

I don't need to find all possible solutions, just any one that satisfies the conditions above.

My question is : is there a better approach other than bruteforcing all possible combinations until I find one? Obviously, greedy algorithm won't work (picking [b, c] for set #1 will make it impossible to pick 2 items from set #2). Do any other options exist? Is my problem equivalent to some well-known problem?

If brute force is the only option, what's the best way to implement it to avoid going down dead-ends? Eg if I pick:

Set #1 [b, e]
Set #2 [a, d]

it would be useless to try all possible combinations for set #3 since picking anyting from set #4 is already impossible.

Answer 1

This can be modelled as a bipartite matching problem : the nodes on the left are the individual elements (a, b, c...) and the nodes on the right are the sets to be chosen from, with multiple nodes representing each set which should be chosen from multiple times. An edge exists from an element (on the left) to a set (on the right) if that element is a member of that set.

The goal is then to compute a matching (ie a set of independent edges) which includes every node on the right; this would necessarily be a maximum cardinality matching, so you can use any algorithm for maximum cardinality bipartite matching . There are several which run in low-degree polynomial time, so these will beat backtracking search for large inputs. (Technically this will be pseudopolynomial , because the transformation to a bipartite graph involves duplicating the sets which should be chosen from multiple times; but the same issue will likely affect backtracking searches.)