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 /***…

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…

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…

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…

暂鸽 #include<cstdio> #include<algorithm> #include<cstring> #include<cmath> #define N 100010 using namespace std; void chkmin(double x,double y) {if (x>y) x=y;} int n; struct point { double x,y; point(){}; point(double _x,double _…

原题地址 题目大意 查询平面内最近点对的距离,输出距离的一半. 暴力做法 枚举每一个点对的距离直接判断,时间复杂度是 $ O(n^2) $,对于这题来说会超时. 那么我们考虑去优化这一个过程,我们在求距离的过程中其实有很多的计算是没有必要的,比如已经有一个暂时的最小值 $ d \(,如果有\) dis>d $,那么这个 $ dis $是没有贡献的,那么我们怎么除去这些不必要的答案呢? 我们可以考虑分治,假设已经求出了两个小区间 $ A , B $的最小值,那么合并的大区间 $ C $ 的最小值实…

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…

Quoit Design Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 25335    Accepted Submission(s): 6716 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…

题意:求平面最近点对之间的距离 解:首先可以想到枚举的方法,枚举i,枚举j算点i和点j之间的距离,时间复杂度O(n2). 如果采用分治的思想,如果我们知道左半边点对答案d1,和右半边点的答案d2,如何求跨两边点之间的答案呢?显然只用枚举中线两边d=min(d1,d2)范围的点,并且每个点都只需要枚举上下范围在d以内的点,显然这样的点不会很多. #include <algorithm> #include <iostream> #include <cstring> #inc…