Ellipsoid

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 658    Accepted Submission(s): 196
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Problem Description Given a 3-dimension ellipsoid(橢球面)

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  

參照了網上的程式，自己寫了一遍。

利用退貨模擬搜尋最優解；
這是一個不斷搜尋最優解的過程，知道最後達到要求的精度的時候不再搜尋了就；
程式碼：

#include <iostream>
#include <math.h>
#include <stdio.h>
#include <string.h>
#define INF 0x3f3f3f3f
#define N 8            
using namespace std;
double a,b,c,d,e,f;
const double r=0.99,eps=1e-9;///r是降溫速度，eps是誤差
int dir[][2]={{0,1},{0,-1},{1,0},{-1,0},{1,1},{1,-1},{-1,1},{-1,-1}};///搜尋的八個方向（也可以是四個）
double dis(double x,double y,double z)
{
    return sqrt(x*x+y*y+z*z);
}
double getz(double x,double y)///根據x，y獲取z的值（二元一次方程）
{
    double A=c;
    double B=d*y+e*x;
    double C=a*x*x+b*y*y+f*x*y-1;
    double detal=B*B-4*A*C;
    if(detal<eps)
        return INF;
    return (sqrt(detal)-B)/(2*A);
}
double slove()
{
    int i;
    double step=1;
    double x=0,y=0,z;
    while(step>eps)///步長開始降溫
    {
        z=getz(x,y);
        for(i=0; i<N; i++)
        {
            double nx=x+step*dir[i][0];
            double ny=y+step*dir[i][1];
            double nz=getz(nx,ny);///獲取下一個x，y，z；
            if(dis(nx,ny,nz)<dis(x,y,z))///找最優解
                x=nx,y=ny,z=nz;
        }
        step*=r;///降溫
    }
    return dis(x,y,z);
}
int main()
{
    double ans;
    while(~scanf("%lf%lf%lf%lf%lf%lf",&a,&b,&c,&d,&e,&f))
    {
        ans=slove();
        printf("%.8lf\n",ans);
    }
    return 0;
}