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哦?难道不是和最近点对一样检查后7个吗?

Posted by wywcgs at 2006-12-25 12:14:00
In Reply To:n个相同的点的时候,不特判的话会退化成O(n^2*logn),至少我的算法现在会这样 Posted by:ArXoR at 2006-12-25 11:59:05
有了周长的p,实际上就只需要检查检查两两距离<p/2的三个点,和最近点对一样的。不还是O(nlgn)么
你是怎么做的?

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