Ellipsoid

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 

1 0.04 0.01 0 0 0

1.0000000

第一次了解模拟退火。

求z时已知x，y转化为关于z的二次方程，用韦达定理求z。

关于退火速度，測了一下，r=0.99时是281ms，r=0.98时是140ms，r=0.97时是93ms，r=0.96时就wa了。

代码：、

//97ms
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
const int dx[8]={0,0,1,-1,1,-1,1,-1};
const int dy[8]={1,-1,0,0,-1,1,1,-1};
const double r=0.97;
const double eps=1e-8;
double a,b,c,d,e,f;
double dis(double x,double y,double z)
{
    return sqrt(x*x+y*y+z*z);
}
double getz(double x,double y)//求z
{
    double A=c;
    double B=d*y+e*x;
    double C=f*x*y+a*x*x+b*y*y-1;
    double delta=B*B-4*A*C;
    if(delta<0)
    {
        return 1e30;
    }
    else
    {
        double z1=(-B+sqrt(delta))/A/2;
        double z2=(-B-sqrt(delta))/A/2;
        return z1*z1<z2*z2?z1:z2;
    }
}
double anneal()//退火
{
    double step=1;
    double x=0,y=0,z;
    while(step>eps)
    {
        z=getz(x,y);
        for(int i=0;i<8;i++)
        {
            double xi=x+dx[i]*step;
            double yi=y+dy[i]*step;
            double zi=getz(xi,yi);
            if(zi>1e20)
            continue;
            if(dis(xi,yi,zi)<dis(x,y,z))
            {
                x=xi;
                y=yi;
                z=zi;
            }
        }
        step=step*r;
    }
    return dis(x,y,z);
}
int main()
{
    while(~scanf("%lf%lf%lf%lf%lf%lf",&a,&b,&c,&d,&e,&f))
    {
       printf("%.8f\n",anneal());
    }
    return 0;
}