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alien dp #7

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beet-aizu opened this issue Sep 11, 2021 · 0 comments
Open

alien dp #7

beet-aizu opened this issue Sep 11, 2021 · 0 comments

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@beet-aizu
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ペナルティ p に対する個数の最大値を cnt(p) とすると、cnt(p) は広義単調減少
p = \argmax_{q} k <= cnt(q) とすると
定義から \all_{0 < eps} cnt(p+eps)<k<=cnt(p)
連続性から lim_{eps->+0} ans(p+eps) = ans(p)
したがって \max_{0 < eps} cnt(p+eps) (<k) 個でもペナルティ p での最適解が存在し、OK
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@beet-aizu