Codeforces Round #299 - Tavas and Pashmaks solution

1. Codeforces Round #299
<Tavas and Pashmaks>
Minsu Kim

2. Description
• Given a set of competitors which have 𝑟𝑖 and 𝑠𝑖 as its value,
find the largest subset of possible winners W
• a competitor can be a winner if there exists 𝑅, 𝑆 ∈ ℝ+
2
where 𝑡 =
𝑅
𝑟 𝑖
+
𝑆
𝑠 𝑖
has the minimum value in the set.
• Constraints : 1 ≤ 𝑛 ≤ 2 × 105, 1 ≤ 𝑠𝑖, 𝑟𝑖 ≤ 104

3. Obvious Facts
• A competitor 𝑖 cannot be a winner if another competitor 𝑗
such that 𝑟𝑖 < 𝑟𝑗 𝑎𝑛𝑑 𝑠𝑖 < 𝑠𝑗 exists.
𝑟
𝑠
(𝑟𝑗, 𝑠𝑗)
(𝑟𝑖, 𝑠𝑖)

4. Geometric Analysis
• 𝑡 should be the minimum for 𝑖 to be a winner
• min 𝑡 =
𝑅
𝑟 𝑖
+
𝑆
𝑠 𝑖
= R, S ∙
1
𝑟 𝑖
,
1
𝑠 𝑖
= R, S × |
1
𝑟 𝑖
,
1
𝑠 𝑖
| × cos 𝜃
• Boxed one is the component of
𝟏
𝒓 𝒊
,
𝟏
𝒔 𝒊
in dir. of 𝐑, 𝐒
• Therefore, suppose R, S is any unit vector in the first
quadrant and find the possible winners!

5. Geometric Analysis
1
𝑟
1
𝑠
(𝑅, 𝑆)
Winners!

6. Geometric Analysis
1
𝑟
1
𝑠
(𝑅, 𝑆)
Winners!

7. Geometric Analysis
1
𝑟
1
𝑠
(𝑅, 𝑆)
Winners!

8. Insight
• Winners turned out to be..
the subset of Convex Hull for the points
1
𝑟 𝑖
,
1
𝑠 𝑖
!
• Exactly, the lower-left part of the Convex Hull.

9. Implementation
• There are some pesky things in implementation..
1. Precision Problem : repeated cross production of fractions
• Use integers.
• 𝑓𝑜𝑟 𝑃
1
𝑝 𝑥
,
1
𝑝 𝑦
, 𝐴
1
𝑎 𝑥
,
1
𝑎 𝑦
, 𝐵
1
𝑏 𝑥
,
1
𝑏 𝑦
, 𝑐ℎ𝑒𝑐𝑘𝑖𝑛𝑔 𝑐𝑜𝑢𝑛𝑡𝑒𝑟 − 𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒 𝑤𝑖𝑙𝑙 𝑏𝑒
• 𝐴 − 𝑃 × 𝐵 − 𝑃 =
(𝑝 𝑥−𝑎 𝑥)(𝑝 𝑦−𝑏 𝑦)
𝑎 𝑥 𝑏 𝑦 𝑝 𝑥 𝑝 𝑦
−
𝑝 𝑦−𝑎 𝑦 𝑝 𝑥−𝑏 𝑥
𝑎 𝑦 𝑏 𝑥 𝑝 𝑥 𝑝 𝑦
> 0
⇔ 𝑎 𝑦 𝑏 𝑥 𝑝 𝑥 − 𝑎 𝑥 𝑝 𝑦 − 𝑏 𝑦 − 𝑎 𝑥 𝑏 𝑦 𝑝 𝑦 − 𝑎 𝑦 𝑝 𝑥 − 𝑏 𝑥 > 0
2. Duplicated Competitors : vary for the Convex Hull algorithm
• Regard as one and check it later