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第一次尝试模拟退火.....

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

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>

using namespace std;

const double eps=1e-8;
const double r=0.99;
const int dir_x[8]={0,0,1,-1,1,-1,1,-1};
const int dir_y[8]={1,-1,0,0,-1,1,1,-1};

double a,b,c,d,e,f;

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

double getZ(double x,double y)
{
	double A=c,B=e*x+d*y,C=a*x*x+b*y*y+f*x*y-1;
	double delta=B*B-4*A*C;
	if(delta<0) return 1e60;
	double z1=(-B+sqrt(delta))/2/A;
	double z2=(-B-sqrt(delta))/2/A;
	if(z1*z1<z2*z2) return z1;
	return z2;
}

double solve()
{
	double step=1;
	double x=0,y=0,z;
	while(step>eps)
	{
		z=getZ(x,y);
		for(int i=0;i<8;i++)
		{
			double nx=x+dir_x[i]*step;
			double ny=y+dir_y[i]*step;
			double nz=getZ(nx,ny);
			if(nz>1e30) continue;
			if(DIST(nx,ny,nz)<DIST(x,y,z))
			{
				x=nx;y=ny;z=nz;
			}
		}
		step=step*r;
	}
	return DIST(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;
}

HDOJ 5017 Ellipsoid

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原文地址：http://blog.csdn.net/ck_boss/article/details/39398513