题目大意:给定平面上的 n 个点,求距离最近的两个点的距离的一半. n <= 10^5.   晕乎乎的度过了一上午... 总之来学习下分治吧233 分治就是把大问题拆成小问题,然后根据对小问题处理出的结果合并成大问题的答案 比如说这道题,如果我们按照X坐标把所有的点分成两组: (木哈哈请叫我盗图狂魔○( ^皿^)っHiahiahia… 像上面我们把点分成了集合S1,S2 点对对应的被划分成3种:S1内,S2内,跨S1.S2 那假若我们分别处理出了S1,S2内的答案,再取个min叫做d 那么跨过分…

这个讲的好: https://phoenixzhao.github.io/%E6%B1%82%E6%9C%80%E8%BF%91%E5%AF%B9%E7%9A%84%E4%B8%89%E7%A7%8D%E6%96%B9%E6%B3%95/ 分治法 先空着 看一下这个第三个方法(随机增量哈希,O(n)) 1.千万不要用unordered_map/hash_map!T飞是肯定的:要手写哈希表,所以码量就很大:手写哈希表方法记一下 2.事实上以d为边长画格子,每次遍历相邻的9个格子,常数要比以d/2边…

思路: 分治 套路题 //By SiriusRen #include <cmath> #include <cstdio> #include <algorithm> using namespace std; ; typedef double db; int n; struct P{db x,y;P(){}P(db X,db Y){x=X,y=Y;}}p[N],t[N]; P operator-(P a,P b){return P(a.x-b.x,a.y-b.y);} db…

意甲冠军:给n坐标点.半一对点之间的距离所需的距离最近. 分析:分而治之的方法,最近点. #include<iostream> #include<algorithm> #include<cmath> using namespace std; #define N 100005 double min(double a,double b) { return a<b? a:b; } struct POINT { double x,y; }; POINT point[N],…

题意是求出所给各点中最近点对的距离的一半(背景忽略). 用分治的思想,先根据各点的横坐标进行排序,以中间的点为界,分别求出左边点集的最小距离和右边点集的最小距离,然后开始合并,分别求左右点集中各点与中间点的距离,从这些距离与点集中的最小距离比较,求得最小距离,此处可按纵坐标排序,将纵坐标距离已经大于之前最小距离的部分都剪枝. 代码如下: #include <bits/stdc++.h> using namespace std; ]; struct point { double x,y; }p[…

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): 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): 29344    Accepted Submission(s): 7688 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…

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…

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…

Quoit Design Time Limit: 5 Seconds      Memory Limit: 32768 KB 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…

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…

题目链接 Quoit Design Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 29547    Accepted Submission(s): 7741 Problem Description Have you ever played quoit in a playground? Quoit is a game in which…

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1107 Quoit Design Time Limit: 5 Seconds      Memory Limit: 32768 KB 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…

主题: Quoit Design Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 136 Accepted Submission(s): 77   Problem Description Have you ever played quoit in a playground? Quoit is a game in which flat rin…

Quoit Design Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 29694    Accepted Submission(s): 7788 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): 28539    Accepted Submission(s): 7469 Problem Description Have you ever played quoit in a playground? Quoit is a game in which flat…

Quoit Design 看懂题意以后发现就是找平面最近点对间距离除以2. 平面上最近点对是经典的分治,我的解析 直接上代码 #include<bits/stdc++.h> using namespace std; #define int long long const int N=1000005; const double inf=1e12; struct node{ double x,y; }p[N]; bool cmp(node a,node b) { if(a.x==b.x)retur…

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…

---恢复内容开始--- Quoit Design Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 47126    Accepted Submission(s): 12323 Problem Description Have you ever played quoit in a playground? Quoit is a game…

Quoit Design Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 30505    Accepted Submission(s): 8017 Problem Description Have you ever played quoit in a playground? Quoit is a game in which flat…

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…

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…

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…

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…

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…