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> #…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3007 Each person had do something foolish along with his or her growth.But,when he or she did this that time,they could not predict that this thing is a mistake and they will want this thing would rather…

Problem Description Each person had do something foolish along with his or her growth.But,when he or she did this that time,they could not predict that this thing is a mistake and they will want this thing would rather not happened.The world king Sco…

题意:给出n个点,求最小包围圆. 解法:这两天一直在学这个神奇的随机增量算法……看了这个http://soft.cs.tsinghua.edu.cn/blog/?q=node/1066之后自己写了好久一直写不对……后来在计算几何的模板上找到了…………orz膜拜一下 代码: #include<stdio.h> #include<iostream> #include<algorithm> #include<string> #include<string.h…

传送门 最小圆覆盖模板. //Achen #include<algorithm> #include<iostream> #include<cstring> #include<cstdlib> #include<vector> #include<cstdio> #include<queue> #include<cmath> ; typedef long long LL; typedef double db; us…

1. 题目描述有n个点,求能覆盖这n个点的半径最小的圆的圆心及半径. 2. 基本思路算法模板http://soft.cs.tsinghua.edu.cn/blog/?q=node/1066定义Di表示相对于P[1]和P[i]组成的最小覆盖圆,如果P[2..i-1]都在这个圆内,那么当前的圆心和半径即为最优解.如果P[j]不在这个圆内,那么P[j]一定在新的最小覆盖圆的边界上即P[1].P[j].P[i]组成的圆.因为三点可以确定一个圆,因此只需要不断的找到不满足的P[j],进而更新最优解即可.其…

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…

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…

今天学习了一下最小圆覆盖, 看了一下午都没看懂, 晚上慢慢的摸索这代码,接合着别人的讲解, 画着图跟着代码一步一步的走着,竟然有些理解了. 最小圆覆盖: 给定n个点, 求出半径最小的圆可以把这些点全部包围, 可以在圆的边界上 下面是我的个人理解. 如果不对, 还请路过大牛指出 先找一个点, 让圆心等于这个点的坐标, 半径等于0, 显然这个对的, 接着找下一个点, 如果只有两个点的话, 那么最小的圆一定是以他们为直径做圆, 接着找第三个点, 如果第三个点在园内或者在边界上, 那么不用更新当前的最小…

Buried memory Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4067    Accepted Submission(s): 2171 Problem Description Each person had do something foolish along with his or her growth.But,when…