hdu 5017 Ellipsoid（模拟退火）

Posted 2020-09-23 wzjhoutai

Ellipsoid

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1140    Accepted Submission(s): 412
Special Judge

Problem Description

Given a 3-dimension ellipsoid(椭球面)

your task is to find the minimal distance between the original point (0,0,0) and points on the ellipsoid. The distance between two points (x₁,y₁,z₁) and (x₂,y₂,z₂) is defined as 

 

Input

There are multiple test cases. Please process till EOF.

For each testcase, one line contains 6 real number a,b,c(0 < a,b,c,< 1),d,e,f(0 ≤ d,e,f < 1), as described above. It is guaranteed that the input data forms a ellipsoid. All numbers are fit in double.

 

Output

For each test contains one line. Describes the minimal distance. Answer will be considered as correct if their absolute error is less than 10^-5.

 

Sample Input

1 0.04 0.01 0 0 0

 

Sample Output

1.0000000

 

Source

2014 ACM/ICPC Asia Regional Xi‘an Online

 

题解：

       这题也是看了看别人的题解。学习了一下模拟退火，感觉不难，挺简单。总体的思路就是8个方向的搜索，退火的意思就是逐渐缩小搜索范围，从而使最后的解足够精确。

#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
using namespace std;
const double sp=0.99,eps=1e-8;
double a,b,c,d,e,f,M=1e9;
double dirx[]={-1,-1,-1,0,0,1,1,1};
double diry[]={-1,0,1,-1,1,-1,0,1};

double dis(double x,double y,double z)
{
    return sqrt(x*x+y*y+z*z);
}

double getz(double x,double y)
{
    double A=0,B=0,C=0;
    A=c;
    B=d*y+e*x;
    C=a*x*x+b*y*y+f*x*y-1;
    double delta=B*B-4*A*C;
    if(delta<0) return M;
    double z1=(sqrt(delta)-B)/(2.0*A),z2=(-sqrt(delta)-B)/(2.0*A);
    if(z1*z1<z2*z2) return z1;
    return z2;
}

double solve()
{
    double x=0,y=0,z=0,tx=0,ty=0,tz=0,step=1;
    z=getz(x,y);
    while(step>eps)
    {
        for(int i=0;i<8;i++)
        {
            tx=x+dirx[i]*step;
            ty=y+diry[i]*step;
            tz=getz(tx,ty);
            if(tz>=M) continue;
            if(dis(tx,ty,tz)<dis(x,y,z))
            {
                x=tx,y=ty,z=tz;
            }
        }
        step*=sp;
    }
    return dis(x,y,z);
}

int main()
{
    while(scanf("%lf%lf%lf%lf%lf%lf",&a,&b,&c,&d,&e,&f)!=EOF)
    {
        printf("%.8lf\n",solve());
    }
    return 0;
}