The description to this problem is so long and so ambiguous, that very few people can understand it.
I myself understood it with 2 hours and made a BIG mistake.
Now, I will explain it to you all.

We don't have to pay much attention to the exact rules, but only the card we can lay.
According to the description, we can only lay skip-turn cards until the last one(of course, the last card can be any one), and make sure that all the continuous cards is connectable. What we must do is to find a card sequence to match this.
BUT, there is some TRAPS here:
1. If the last-laid card is QUEEN, then is must be followed by a same-suit card, it means that, for example, we can't use a QC to cover a QD.
2. Jokers may be the most useful cards, but it is the hardest part to this problem(at least I think so). To use it, you can of course assign its value and suit arbitrary, but please make sure that it can match its previous and next card. For instance, you can't use joker here: 3C * JH,
for * can't be a skip-turn card here.

That's all. My solution is not very good, it got ACCEPTED in 0.156 sec.
May all of you a better result than me.

Edited by author 18.08.2008 11:03

I think we can use joker here: 3C * JH,
when wei set * as 8QCH.
(We can do this,see the sample input)

Edited by author 08.02.2009 17:18

Edited by author 08.02.2009 17:18

Now I know I am wrong!
You can only use 6,7,A and KS.(Except the last card).
Thanks to 198808xc !
Now I got AC!!

Edited by author 10.02.2009 12:50

Edited by author 10.02.2009 18:24

But skipping card may be any one,  9H , for example,
if second player hasn't connectable card for it
and upper card in desk also isn't connectable.

Edited by author 14.06.2009 09:50

But, second player may has Joker. Hovewer, 9H - never skipping card. Am I right?


AC~!!! 0.031 in tje first attempt

Edited by author 16.02.2012 21:31