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…

题意: 给出平面上的n个点,问任意点对之间的最短距离是多少? 思路: 先将所有点按照x坐标排序,用二分法将n个点一分为二个部分,递归下去直到剩下两或一个点.对于一个部分,左右部分的答案分别都知道,那么可能的答案就是min(left_ans,right_ans) .注意更小的点对可能一个在左,一个在右.所以还得处理两个边内的紧靠着的部分,如果左边的一个点到达中线的距离已经超过当前最短距离,那么这个点到达右边任意一个点也不会是最短距离了.同时,若一左一右的两个点的y距离已经超过目前最短距离,那么也不…

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…

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): 47104    Accepted Submission(s): 12318 Problem Description Have you ever played quoit in a playground? Quoit is a game in which fla…

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…

大致题意:给N个点,求最近点对的距离 d :输出:r = d/2. // Time 2093 ms; Memory 1812 K #include<iostream> #include<cstdio> #include<cmath> #include<algorithm> #define eps 1e-8 #define maxn 100010 #define sqr(a) ((a)*(a)) using namespace std; int sig(dou…

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 /***…