模拟退火算法

模拟退火算法介绍

1.1 模拟退火算法的原理

模拟退火算法是一种元启发式（Meta-Heuristics）算法，来源于固体退火原理，将固体加热至充分高的温度，再让其徐徐冷却。加热时，固体内部粒子随温升变为无序状，内能增大，而徐徐冷却时粒子渐趋有序，在每个温度都达到平衡态，最后在常温时达到基态，内能减为最小。根据Metropolis准则，粒子在温度T时趋于平衡的概率为 ，其中E为温度T时的内能，ΔE为其改变量，k为Boltzmann常数。

1.2 模拟退火算法的模型

① 初始化：初始温度T(充分大)，初始解状态S(算法迭代的起点)， 每次迭代次数L

② for k=1 to L 做③至⑥

③ 产生新解S’

④ 计算增量Δt′=C(S′)-C(S)，其中C(S)为评价函数

⑤ 若Δt′<0则接受S’作为新的当前解，否则以概率 接受S’作为新的当前解

⑥ 如果满足终止条件则输出当前解作为最优解，结束程序

⑦ T逐渐减少，然后转②

 

又是一种神奇的算法：模拟退火。就是一步步地退火，一步步地逼进。这种算法应该是只能用于一些单调函数性质的几何问题、现实问题吧。模拟退火与以往所学的算法大不相同，在某种程度上来说是一种万能算法。如果有什么图论、几何之类的题如果做不出，应该可以用模拟退火来试试人品。

POJ 2420 照着网上一个小代码轻松敲定。人品不错，一次AC。

Source Code
Problem: 2420 User: liu696639Memory: 352K Time: 0MSLanguage: C++ Result: Accepted
Source Code
#include<iostream>#include<cmath>#include<stdio.h>using namespace std;struct POINT {       double x,y;       POINT(){x=0;y=0;}       POINT (double a,double b){x=a;y=b;}}point[10000];double dist(POINT a,POINT b){       return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));}int n;double all(POINT b){       double sum=0;       for(int i=0;i<n;i++)       {               sum+=dist(b,point[i]);       }       return sum;}int dir[4][2]={0,1,0,-1,1,0,-1,0};//int dir[4][2]={(0,1),(0,-1),(1,0),(-1,0)};int main(){    while( scanf("%d",&n)!=EOF && n!=0)    {            for( int i=0;i<n;i++)            {                 scanf("%lf%lf",&point[i].x,&point[i].y);            }            POINT ans=point[0];            double ret=all(ans);            POINT temp;            double step=500000;                        while( step>0.0002)            {                   int flag=1;                   while( flag)                   {                            temp=ans;                            flag=0;                            for(int i=0;i<4;i++)                            {                                   temp=POINT(ans.x+step*dir[i][0],ans.y+step*dir[i][1]);                                   double tempdist=all(temp);                                   if(ret>tempdist)                                   {                                        flag=1;                                        ans=temp;                                        ret=tempdist;                                        break;                                   }//cout<<ret<<"  ";                            }                   }                   step/=2.0;                               }//cout<<endl;            printf("%.lf\n",ret);    }}

poj 2069 SUPER STAR:求点集的最小外接球

WA很多次，最后还是照着网上代码过的。自己一个人处理精度与寻找评估函数的能力还是很欠缺。

Problem: 2069User: liu696639Memory: 200KTime: 16MSLanguage: C++Result: Accepted

Source Code

#include<iostream>
#include<cmath>
#include<stdio.h>
using namespace std;
const double EP=1E-10;
const int maxn=10000;
struct POINT 
{
       double x,y,z;
       POINT(double a,double b,double c){x=a,y=b,z=c;}
       POINT(){}
}point[maxn];
double dist(POINT a,POINT b)
{
       return sqrt( (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)+(a.z-b.z)*(a.z-b.z) );
}
int n;
int maxdist(POINT p)
{
       double ret=0,l;
       int k=0;
       for(int i=0;i<n;i++)
       {
               
               l=dist(p,point[i]);
               if(l>ret) 
               {
                         ret=l;
                         k=i;
               }
       }
       return k;
}
int main()
{
    
    while( scanf("%d",&n)!=EOF && n!=0)
    {
           for(int i=0;i<n;i++)
           {
                   scanf("%lf %lf %lf",&point[i].x,&point[i].y,&point[i].z);
           }
           POINT p(0,0,0);
           double step=100;
           double ans=dist(point[maxdist(p)],p);
           int t=0;
           while( step>=EP)
           {
                  t++;
                  int id=maxdist(p);
                  double temp=dist(p,point[id]);
                  if(temp<ans)
                  ans=temp;
                  double dx=point[id].x-p.x;
                  double dy=point[id].y-p.y;
                  double dz=point[id].z-p.z;
                  p.x=p.x+(dx/temp)*step;
                  p.y=p.y+(dy/temp)*step;
                  p.z=p.z+(dz/temp)*step;
                  step*=0.99;
           }
           printf("%.5lf\n",ans);
    }
}

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