I'm trying to solve this problem on LightOJ. What i've done is to calculate the total perimeter of the hull points and then add the additional curved portion as (pi-theta)*r [from picture below]
But i'm getting wrong answer. Is my approach wrong?
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I'm trying to solve this problem on LightOJ. What i've done is to calculate the total perimeter of the hull points and then add the additional curved portion as (pi-theta)*r [from picture below]
But i'm getting wrong answer. Is my approach wrong?
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I think the solution is the perimeter of Convex Hull + perimeter of the circunference of ratio d. Because all curved portions form a circunference. The proof is easy, only show that the sum of all theta is 2*pi.