bzoj3680 -- 模拟退火

模拟退火裸题（输出"nan nan"可以AC）

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
using namespace std;
#define N 10010
struct Node{
    double x,y,z;
}a[N];
double x,y,T=100000,Now,Ax,Ay,X,Y,N2,Min=1e17;
int i,j,k,n,m;
inline double Dist(double x,double y,Node a){
    return sqrt((x-a.x)*(x-a.x)+(y-a.y)*(y-a.y));
}
inline double Calc(double X,double Y){
    double Ans=0;
    for(int i=1;i<=n;i++)Ans+=Dist(X,Y,a[i])*a[i].z;
    if(Ans<Min)Ax=X,Ay=Y,Min=Ans;
    return Ans;
}
inline double Getrand(){
    return rand()%1000/1000.0;
}
inline void SA(){
    while(T>0.001){
        X=x+T*(Getrand()*2-1);Y=y+T*(Getrand()*2-1);
        N2=Calc(x,y)-Calc(X,Y);
        if(N2>0||exp(N2/T)>Getrand())x=X,y=Y;
        T*=0.993;
    }
    for(int i=1;i<=1000;i++){
        X=Ax+T*(Getrand()*2-1);Y=Ay+T*(Getrand()*2-1);Calc(X,Y);
    }
}
int main(){
    srand(10086);
    scanf("%d",&n);
    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;
    x/=n;y/=n;
    SA();
    printf("%.3lf %.3lf
",Ax,Ay);
    return 0;
}