HDU 5017 ellipsoid (Simulated Annealing)

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 descriptiongiven a 3-dimension ellipsoid (elliptical)

Your task is to find the minimal distance between the original point (0, 0) and points on the ellipsoid. the distance between two points (x1, Y1, Z1) and (X2, Y2, Z2) is defined
Inputthere 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 abve. It is guaranteed that the input data forms a ellipsoid.All numbers are fit in double.
 
Outputfor 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

 
Source2014 ACM/ICPC Asia Regional Xi 'an online
Question: I also looked at others' questions and learned simulated annealing. It is not difficult and simple. The overall idea is to search in eight directions. The annealing means gradually narrowing down the search scope so that the final solution is accurate enough.

#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;}

HDU 5017 ellipsoid (Simulated Annealing)