Quoit Design(hdu1007)最近点对问题。模版哦！

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Quoit Design(hdu1007)最近点对问题。模版哦！

Quoit Design Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 30919 Accepted Submission(s): 8120 Problem Description Have you ever played quoit in a playground? Quoit is a game in which flat rings…

Quoit Design（最近点对+分治）

题目链接: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): 42865 Accepted Submission(s): 11128 Problem Description Have you ever played…

ACM-计算几何之Quoit Design——hdu1007 zoj2107

Quoit Design Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 28539 Accepted Submission(s): 7469 Problem Description Have you ever played quoit in a playground? Quoit is a game in which flat…

HDU-1007 Quoit Design 平面最近点对

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1007 简单裸题,测测模板,G++速度快了不少,应该是编译的时候对比C++优化了不少.. //STATUS:G++_AC_1703MS_5004KB #include <functional> #include <algorithm> #include <iostream> //#include <ext/rope> #include <fstream>…

HDOJ-1007 Quoit Design（最近点对问题）

http://acm.hdu.edu.cn/showproblem.php?pid=1007 给出n个玩具(抽象为点)的坐标求套圈的半径要求最多只能套到一个玩具实际就是要求最近的两个坐标的距离典型的最近点对问题最近点对详解 http://blog.csdn.net/lonelycatcher/article/details/7973046 //最近点对 # include <stdio.h> # include <algorithm> # include <math…

【HDOJ】P1007 Quoit Design （最近点对）

题目意思很简单,意思就是求一个图上最近点对. 具体思想就是二分法,这里就不做介绍,相信大家都会明白的,在这里我说明一下如何进行拼合. 具体证明一下为什么只需要检查6个点首先,假设当前左侧和右侧的最小值为d,那么对于一个点,如果有个最小值小于d,那么一定存在于上d下d左d右d的一块区域内,又因为是从左到右,从上到下,所以左侧的那部分匹配的区域会重叠,也就是对于左侧的区域,完全没有必要去进行匹配.所以只需要对右侧d,上下d的区域进行匹配,而假设这个区域内的所以点的距离为d那么最多为6个,那么如果长…

杭电OJ——1007 Quoit Design(最近点对问题)

Quoit Design Problem Description Have you ever played quoit in a playground? Quoit is a game in which flat rings are pitched at some toys, with all the toys encircled awarded. In the field of Cyberground, the position of each toy is fixed, and the ri…