模拟退火

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题目链接
题意：给出椭球的6个参数，求椭球上一点，到原点的距离最短。
别人讲的模拟退火

#include<cstdio>#include<cstring>#include<cmath>#include<vector>#include<iostream>#include<algorithm>using namespace std;const double inf=0x3f3f3f3f;const int maxn=1e4+10;const double eps=1e-8;const double cooling=0.99;//降温速度double a,b,c,d,e,f;int dx[10]={0,0,1,1,1,-1,-1,-1};int dy[10]={1,-1,0,-1,1,0,-1,1};double get_dis(double x,double y,double z)//计算到原点的距离{    return sqrt(x*x+y*y+z*z);}double get_z(double x,double y)//根据x和y计算z的值{    double C=a*x*x+b*y*y+f*x*y-1,B=e*x+d*y,A=c;    double s=B*B-4*A*C;    if(s<0) return inf;    double z1=(0-B+sqrt(s))/(2.0*A);    double z2=(0-B-sqrt(s))/(2.0*A);    return fabs(z1)>fabs(z2)?z2:z1;}double solve()//模拟退火{    double step=1;//步长    double x=0,y=0,z;    while(step>eps)    {        z=get_z(x,y);        for(int i=0;i<8;i++)//8个方向走        {            double nx=x+dx[i]*step;            double ny=y+dy[i]*step;            double nz=get_z(nx,ny);            if(get_dis(x,y,z)>get_dis(nx,ny,nz))//找到更优解就更新                x=nx,y=ny,z=nz;        }        step*=cooling;//退火    }    return get_dis(x,y,z);//返回最优解}int main(){    while(~scanf("%lf%lf%lf%lf%lf%lf",&a,&b,&c,&d,&e,&f))    {        printf("%.7lf\n",solve());    }    return 0;}

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