HDOJ-1007 Quoit Design（最近点对问题）

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最近点对问题 POJ 3714 Raid && HDOJ 1007 Quoit Design

题意:有n个点,问其中某一对点的距离最小是多少分析:分治法解决问题:先按照x坐标排序,求解(left, mid)和(mid+1, right)范围的最小值,然后类似区间合并,分离mid左右的点也求最小值 POJ 3714 #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> const int N = 1e5 + 5; const double INF = 1e…

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…

hdu 1007 Quoit Design (最近点对问题)

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…

Hdoj 1007 Quoit Design 题解

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 careful…

HDU 1007 Quoit Design最近点对（分治法）

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…