Thank you,I got AC!

My solution is based on the concept of strong connectivity. I find connected components of the graph, then I print those jedis, that belong to the non-dominated component. It is npt very quick, but it works. I don't know, how BFS can help with this problem. Any explanation will be very much appreciated.

(de bene esse: my e-mail is akhmed[at]astranet[dot]ru).

for each start among knights
 mark all that can be reached from start thru winning
 if all are marked then print start's name

To test your speed, you can use the following perl script to generate input:

$n = 200;
print "$n\n";
for(1..$n){
    print "J$_"; print ' ', int(rand()*100000) for 1..3; print "\n"
}

O(N*log(N)) solution exists :)