Quoit Design Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 48729    Accepted Submission(s): 12823 Problem Description Have you ever played quoit in a playground? Quoit is a game in which flat…

Quoit Design Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 47104    Accepted Submission(s): 12318 Problem Description Have you ever played quoit in a playground? Quoit is a game in which fla…

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 ring is carefull…

Quoit Design Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 58566    Accepted Submission(s): 15511 Problem Description Have you ever played quoit in a playground? Quoit is a game in which flat…

http://acm.hdu.edu.cn/showproblem.php?pid=1007 最近欧式距离模板题. 用分治大法(分治的函数名用cdq纯属个人习惯_(:з」∠)_) 一开始狂M. 后来判断n是否为0就不M了QwQ #include<cmath> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; const int N = 100003; str…

Shape of HDU Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4972    Accepted Submission(s): 2250 Problem Description 话说上回讲到海东集团推选老总的事情,最终的结果是XHD以微弱优势当选,从此以后,“徐队”的称呼逐渐被“徐总”所取代,海东集团(HDU)也算是名副其实了.…

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…

[题目链接] http://acm.hdu.edu.cn/showproblem.php?pid=1007 [算法] 答案为平面最近点对距离除以2 [代码] #include <algorithm> #include <bitset> #include <cctype> #include <cerrno> #include <clocale> #include <cmath> #include <complex> #inc…

http://acm.hdu.edu.cn/showproblem.php?pid=1007 上半年在人人上看到过这个题,当时就知道用分治但是没有仔细想... 今年多校又出了这个...于是学习了一下平面内求最近点对的算法...算导上也给了详细的说明 虽然一看就知道直接用分治O(nlogn)的算法 , 但是平面内最近点对的算法复杂度证明我看了一天也没有完全看明白... 代码我已经做了一些优化...但肯定还能进一步优化..我是2s漂过的非常惭愧...(甚至优化以后时间还多了...不明白原因 /***…

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…