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Discussion of Problem 1218. Episode N-th: The Jedi Tournament

Dmitry 'Diman_YES' Kovalioff BFS (-) [3] // Problem 1218. Episode N-th: The Jedi Tournament 28 Jul 2004 09:27
Saturn Re: BFS (-) // Problem 1218. Episode N-th: The Jedi Tournament 28 Jul 2004 12:06
Thank you,I got AC!
UXMRI: Sergey Baskakov, Raphail Akhmedisheff and Denis Nikonorov I think I used a different approach. Please, specify yours more precisely. [1] // Problem 1218. Episode N-th: The Jedi Tournament 21 Nov 2005 00:44
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"
}
Alexey Dergunov [Samara SAU] Re: How to solve this problem? // Problem 1218. Episode N-th: The Jedi Tournament 8 Jul 2012 03:35
O(N*log(N)) solution exists :)