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Slightly? Harder version of the Div2-500/Div1-250 | Reply
Same setup, but now the question is not who will win, since we already know.

The question is now this: Knowing who will win, what is the least number of moves in which they can end the game.

The player who can win will always aim for the least number of moves, while the player who can't win will always move so as to make the game last as long as possible.

An example:

The game board looks like this:

In this game, player 2 will win, but player 1 goes first.
Player 1 has two options, but he will choose --> oo.. rather than .oo.

So the answer was '6' to this board.


Ans: 8
There can be more than one way to attain the minimum game-length.
Hah, another variation could be where the losing player assists the winning player in ending the game as soon as possible.
Re: Slightly? Harder version of the Div2-500/Div1-250 (response to post by Cricicle) | Reply
The "slightly" harder version of this problem was a 2D board up to 50 by 50, which was actually the inInternal Error

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