zoukankan      html  css  js  c++  java
  • poj 2069 Super Star

    Super Star
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 5752   Accepted: 1433   Special Judge

    Description

    During a voyage of the starship Hakodate-maru (see Problem 1406), researchers found strange synchronized movements of stars. Having heard these observations, Dr. Extreme proposed a theory of "super stars". Do not take this term as a description of actors or singers. It is a revolutionary theory in astronomy. 
    According to this theory, starts we are observing are not independent objects, but only small portions of larger objects called super stars. A super star is filled with invisible (or transparent) material, and only a number of points inside or on its surface shine. These points are observed as stars by us. 

    In order to verify this theory, Dr. Extreme wants to build motion equations of super stars and to compare the solutions of these equations with observed movements of stars. As the first step, he assumes that a super star is sphere-shaped, and has the smallest possible radius such that the sphere contains all given stars in or on it. This assumption makes it possible to estimate the volume of a super star, and thus its mass (the density of the invisible material is known). 

    You are asked to help Dr. Extreme by writing a program which, given the locations of a number of stars, finds the smallest sphere containing all of them in or on it. In this computation, you should ignore the sizes of stars. In other words, a star should be regarded as a point. You may assume the universe is a Euclidean space. 

    Input

    The input consists of multiple data sets. Each data set is given in the following format. 


    x1 y1 z1 
    x2 y2 z2 
    . . . 
    xn yn zn 

    The first line of a data set contains an integer n, which is the number of points. It satisfies the condition 4 <= n <= 30. 

    The location of n points are given by three-dimensional orthogonal coordinates: (xi, yi, zi) (i = 1, ..., n). Three coordinates of a point appear in a line, separated by a space character. Each value is given by a decimal fraction, and is between 0.0 and 100.0 (both ends inclusive). Points are at least 0.01 distant from each other. 

    The end of the input is indicated by a line containing a zero. 

    Output

    For each data set, the radius of the smallest sphere containing all given points should be printed, each in a separate line. The printed values should have 5 digits after the decimal point. They may not have an error greater than 0.00001.

    Sample Input

    4
    10.00000 10.00000 10.00000
    20.00000 10.00000 10.00000
    20.00000 20.00000 10.00000
    10.00000 20.00000 10.00000
    4
    10.00000 10.00000 10.00000
    10.00000 50.00000 50.00000
    50.00000 10.00000 50.00000
    50.00000 50.00000 10.00000
    0

    Sample Output

    7.07107
    34.64102
    

    Source

     
    题意:给定三维空间的n个点,找出一个半径最小的球把这些点全部包围住。
     
    模拟退火
    每次中心点向距当前中心点最远的点距离移动,退火把握移动距离
     
    #include<cmath>
    #include<cstdio>
    #include<algorithm>
    
    #define N 31
    
    using namespace std;
    
    #define eps 1e-8
    #define delta 0.98
    #define T 100
    
    int n;
    struct Point
    {
        double x,y,z;
    };
    Point P[N];
    double ans;
    
    double dist(Point A,Point B)
    {
        return sqrt((A.x-B.x)*(A.x-B.x)+(A.y-B.y)*(A.y-B.y)+(A.z-B.z)*(A.z-B.z));
    }
    
    double solve()
    {
        double t=T;
        Point s=P[0];
        int k; double mx,res;
        ans=1e20;
        while(t>eps)
        {
            mx=0;
            for(int i=1;i<=n;i++)
            {
                res=dist(s,P[i]);
                if(res>mx) mx=res,k=i;
            }
            ans=min(ans,mx);
            s.x+=(P[k].x-s.x)/mx*t;
            s.y+=(P[k].y-s.y)/mx*t;
            s.z+=(P[k].z-s.z)/mx*t;
            t*=delta;
        }
        return ans;
    }
    
    int main()
    {
        while(scanf("%d",&n)!=EOF)
        {
            if(!n) return 0;
            for(int i=1;i<=n;i++) scanf("%lf%lf%lf",&P[i].x,&P[i].y,&P[i].z);
            printf("%.5lf
    ",solve());
        }    
    }
  • 相关阅读:
    ble_app_hrs心率程序 nrf51822
    2019.05.08 《Linux驱动开发入门与实战》
    函数指针
    typedef
    回调函数
    android2
    android1
    每周总结2
    HTML
    数组(续)
  • 原文地址:https://www.cnblogs.com/TheRoadToTheGold/p/7645931.html
Copyright © 2011-2022 走看看