Notice that there are 18*3=54 participants, 64bit-type should be used for masks.

It's possible to solve that problem with only 32-bit masks. Encode team number k as  2^k (in c "1 << k", in pascal "1 shl k").

Create an array (I named it "check"):

forall (i)
{
  check[i] = 0;
  forall (j, not equal i)
  {
    if (there is a member in team j that exists also in team i)
      then check[i] = check[i] bitwise or (2 ^ j);
  }
}

You can encode every possible teams combination in one number from interval [0 .. (2^k) - 1]. Note that the number is less than 2 ^ 18 (or 262144).

Then enumerate all numbers from that interval and for every number S for every bit p in position q in number S check inequality "S & check[q] > 0".

If for some p in position q in number S this inequality is true, than state S is inacceptable.

Edited by author 23.09.2011 01:18