Hi all.
I use dfs to find the number of connected components.
If there are 2 components,then i output each of them as a single team.If there is only 1 CC,then i output numbers from 1 to n/2  and from n/2+1 to n.
Otherwise "No solution".
It seems to me correct,but....
Or tell me some tests that prove my idea's fail)

Your idea is incorrect.

1) If 1st person knows 2nd person it doesn't mean that 2nd person knows 1st person.
2) If 1st and 2nd persons know each other and 1st and 3rd persons also know each other it doesn't mean that 2nd person knows 3rd person and vice versa.

According to your algo in the first case both persons will be in one connected component, and in the second case all 3 persons will be in one connected component.

Hi, I think that you should find the strongest connected components in a digraph.