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?
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.