Maple trees Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 222 Accepted Submission(s): 79   Problem Description There are a lot of trees in HDU. Kiki want to surround all the trees with the minim…

称号: Maple trees Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 177 Accepted Submission(s): 63   Problem Description There are a lot of trees in HDU. Kiki want to surround all the trees with the m…

Maple trees Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1847    Accepted Submission(s): 574 Problem Description There are a lot of trees in HDU. Kiki want to surround all the trees with the…

题目链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=450 You are to write a program to find a circle which covers a set of points and has the minimal area. There will be no more than 100 points in one problem. 题意描述:找到一个最小圆能够包含到所有的二维坐标点. 算法…

大致题意: 平面上有n个点,求一个最小的圆覆盖住所有点 最小覆盖圆裸题 学习了一波最小覆盖圆算法 #include<cstdio> #include<iostream> #include<cstring> #include<algorithm> #include<queue> #include<set> #include<map> #include<stack> #include<time.h> #…

增量法的最小包围圈算法,不会…… #include <cstdio> #include <cstring> #include <iostream> #include <cmath> #include <algorithm> using namespace std; const double EPS = 1e-10; inline int sgn(double x) { return (x > EPS) - (x < -EPS);} s…

clear all; close all; clc; n=100; p=rand(n,2); p1=p(1,:); %取第一行的值 P1点 p2=p(2,:); %取第二行的值 P2点 r=sqrt((p1(1)-p2(1))^2+(p1(2)-p2(2))^2)/2; %求两点半径 cenp=(p1+p2)/2; %求两点中点 for i=3:n newp=p(i,:); %从第三行开始 储存新的点 d=sqrt((cenp(1)-newp(1))^2+(cenp(2)-newp(2))^2)…

最小覆盖圆算法.看着题解半蒙半抄的搞过去了… 主要参考以下http://blog.csdn.net/acdreamers/article/details/9406735http://blog.csdn.net/lthyxy/article/details/6661250http://blog.himdd.com/archives/2666 其中有一个求外心的过程是错的…害我调了好久…还把x和y搞反…可是就算是错的我居然过了80%的数据… 仍然几个疑问: (1)最小覆盖圆算法时间复杂度的证明 (2…

首先通过随机增量法求出最小覆盖圆,作为答案的上界. 然后二分答案,检验的时候枚举每个点作为原点,求出其他每个点被包括在圆内的角度区间,然后扫描线即可. 时间复杂度$O(Tn^2\log n)$. #include<cstdio> #include<cmath> #include<algorithm> #include<cstdlib> #define N 510 using namespace std; const double PI=acos(-1.0),…

题意 给\(n\)个点,求一个能覆盖所有点的面积最小的圆.(\(n \le 50000\)) 分析 随机增量法 题解 理论上\(O(n^3)\)暴力,实际上加上随机化后期望是\(O(n)\)的. 算法大概就是: 假设我们已经得到了最小覆盖圆\(O\),然后现在考虑假如第\(i\)个点进去. 如果第\(i\)个点在圆内或在圆上,则不需要更改.如果在圆外,显然最小覆盖圆要经过这个点. 于是又从头考虑\(1 \sim i-1\)这些点,我们只需要找到一个经过\(i\)点的覆盖所有点的最小覆盖圆.于是同…