bzoj3680 -- 模拟退火

来源：互联网发布：淘宝台湾省编辑：程序博客网时间：2024/05/22 14:00

模拟退火裸题（输出"nan nan"可以AC）

代码：

 1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cmath> 5 #include<cstdlib> 6 using namespace std; 7 #define N 10010 8 struct Node{ 9     double x,y,z;10 }a[N];11 double x,y,T=100000,Now,Ax,Ay,X,Y,N2,Min=1e17;12 int i,j,k,n,m;13 inline double Dist(double x,double y,Node a){14     return sqrt((x-a.x)*(x-a.x)+(y-a.y)*(y-a.y));15 }16 inline double Calc(double X,double Y){17     double Ans=0;18     for(int i=1;i<=n;i++)Ans+=Dist(X,Y,a[i])*a[i].z;19     if(Ans<Min)Ax=X,Ay=Y,Min=Ans;20     return Ans;21 }22 inline double Getrand(){23     return rand()%1000/1000.0;24 }25 inline void SA(){26     while(T>0.001){27         X=x+T*(Getrand()*2-1);Y=y+T*(Getrand()*2-1);28         N2=Calc(x,y)-Calc(X,Y);29         if(N2>0||exp(N2/T)>Getrand())x=X,y=Y;30         T*=0.993;31     }32     for(int i=1;i<=1000;i++){33         X=Ax+T*(Getrand()*2-1);Y=Ay+T*(Getrand()*2-1);Calc(X,Y);34     }35 }36 int main(){37     srand(10086);38     scanf("%d",&n);39     for(i=1;i<=n;i++)scanf("%lf%lf%lf",&a[i].x,&a[i].y,&a[i].z),x+=a[i].x,y+=a[i].y;40     x/=n;y/=n;41     SA();42     printf("%.3lf %.3lf\n",Ax,Ay);43     return 0;44 }

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