hdu 5017 Ellipsoid（模擬退火）

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(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

2014 ACM/ICPC Asia Regional Xi'an Online

 

題解：

       這題也是看了看別人的題解，學習了一下模擬退火，感覺不難，挺簡單。整體的思路就是8個方向的搜索，退火的意思就是逐漸縮小搜索範圍，從而使最後的解足夠精確。

#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;
}