(hdu 7.1.8)Quoit Design(最低点——在n一个点,发现两点之间的最小距离)

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(hdu 7.1.8)Quoit Design(最低点——在n一个点,发现两点之间的最小距离)

主题: Quoit Design Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 136 Accepted Submission(s): 77 Problem Description Have you ever played quoit in a playground? Quoit is a game in which flat rin…

HDU 1007 Quoit Design（二分+浮点数精度控制）

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HDU 1007 Quoit Design（经典最近点对问题）

传送门: http://acm.hdu.edu.cn/showproblem.php?pid=1007 Quoit Design Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 62916 Accepted Submission(s): 16609 Problem Description Have you ever played…

hdu 1007 Quoit Design (最近点对问题)

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HDU 1007 Quoit Design【计算几何/分治/最近点对】

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hdu 1007 Quoit Design 分治求最近点对

Quoit Design Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 29344 Accepted Submission(s): 7688 Problem Description Have you ever played quoit in a playground? Quoit is a game in which flat…

HDU 1007：Quoit Design（分治求最近点对）

http://acm.hdu.edu.cn/showproblem.php?pid=1007 题意:平面上有n个点,问最近的两个点之间的距离的一半是多少. 思路:用分治做.把整体分为左右两个部分,那么有三种情况:最近的两个点都在左边,最近的两个点都在右边和最近的两个点一个在左边一个在右边.对于第一第二种情况,直接递归处理,分解成子问题就好了,主要是要处理第三种情况.最暴力的做法是O(n^2)的扫,这样肯定会TLE.那么要作一些优化.首先我们先递归处理得到第一种和第二种情况的答案的较小值,然后用这…