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Обсуждение задачи 1520. Империя наносит ответный удар

Страницы: Предыдущая 1 2
Re: Probability will rule?
Послано Vedernikoff Sergey 18 янв 2007 16:46
If you drop only 50 points, you'll not detect many little "white spaces" (when, for example, current radius 1e-3 less, than optimal. Obviously, such a solution will get WA.
How to check coverage with bruteforce? Intersect any two circles and consider epsilon neighborhood of their common points...
Re: Probability will rule?
Послано SPIRiT 18 янв 2007 17:21
Thanks a lot. But perhaps if I increase the number of random points (for each check a separate set of points), I'll get AC (if not from the first time, than from the second thanks to the random :)
Re: Probability will rule?
Послано Vedernikoff Sergey 19 янв 2007 14:30
To my mind, it's practically impossible. You should check (1000/1e-5)^2 ~ 10^16 points on every step of binsearch to catch "little whitespaces". Quadrotrees can help you a little, but...
Re: Probability will rule?
Послано KIRILL(ArcSTU) 19 янв 2007 14:46
Vedernikoff Sergey писал(a) 18 января 2007 16:46
If you drop only 50 points, you'll not detect many little "white spaces" (when, for example, current radius 1e-3 less, than optimal. Obviously, such a solution will get WA.
How to check coverage with bruteforce? Intersect any two circles and consider epsilon neighborhood of their common points...

Your idea is very good, but how you did it so fast?
I have TL16
Re: Can you give me a hint on how to solve this?
Послано Yogesh 20 янв 2007 22:02
theriyaathutaa koothi
Re: Probability will rule?
Послано Vedernikoff Sergey 22 янв 2007 14:55
Apply some optimixation techniques - precalc, for instance...
Re: Probability will rule?
Послано KIRILL(ArcSTU) 22 янв 2007 19:13
Vedernikoff Sergey писал(a) 22 января 2007 14:55
Apply some optimixation techniques - precalc, for instance...

Now I have WA17
I do not understand what points should we check
If we will check only neighboor points of
intersection any 2 circles - it's not right

For example
2 1000
1000 0
999 0

Right answer is 1999

But if we will check radius 1000 then both intersection points will lie out of main circle
Re: Probability will rule?
Послано it4.kp 22 янв 2007 22:25
For each two crossing circles (crossing inside main circle) there must be third one which covers intersection point of the first two circles.
Re: Probability will rule?
Послано nttjuvwamsncc 14 фев 2007 14:33
can you tell me more about your solution
Re: Can you give me a hint on how to solve this?
Послано nttjuvwamsncc 14 фев 2007 14:33
how can you solve this problem without the diagram
Re: Probability will rule?
Послано nttjuvwamsncc 14 фев 2007 14:35
I don't understand you solution
tell me more
Страницы: Предыдущая 1 2