题意：平面上找一个点，使得其到给定的n个点的距离的最小值最大。

模拟退火看这篇：http://www.cnblogs.com/autsky-jadek/p/7524208.html

这题稍有不同之处仅有：多随机几个初始点，以增加正确率。

另：WA了几百遍竟然是因为最后输出了-0.0这样的值……

#include
     
      #include
      
       #include
       
        #include
        
         using namespace std;const double EPS=0.00000001;const double PI=acos(-1.0);struct Point{	double x,y;	Point(const double &x,const double &y){		this->x=x;		this->y=y;	}	Point(){}	void read(){		scanf("%lf%lf",&x,&y);	}	double length(){		return sqrt(x*x+y*y);	}}a[1005],p,allp;double ans,allans;int n,X,Y;typedef Point Vector;Vector operator - (const Point &a,const Point &b){	return Vector(a.x-b.x,a.y-b.y);}Vector operator + (const Vector &a,const Vector &b){	return Vector(a.x+b.x,a.y+b.y);}Vector operator * (const double &K,const Vector &v){	return Vector(K*v.x,K*v.y);}double calc(Point p){	double res=1000000.0;	for(int i=1;i<=n;++i){		res=min(res,(a[i]-p).length());	}	return res;}int main(){	int zu;	srand(233);	//freopen("poj1379.in","r",stdin);	//freopen("poj1379.out","w",stdout);	scanf("%d",&zu);	for(;zu;--zu){		allans=0.0;		scanf("%d%d%d",&X,&Y,&n);		p=Point(0.0,0.0);		for(int i=1;i<=n;++i){			a[i].read();		}		for(int j=1;j<=15;++j){			p.x=(double)(rand()%(X+1));			p.y=(double)(rand()%(Y+1));			ans=calc(p);			double T=sqrt((double)X*(double)X+(double)Y*(double)Y)/2.0;//			double T=(double)max(X,Y)/sqrt((double)n);			while(T>EPS){				double bestnow=0.0;				Point besttp;				for(int i=1;i<=35;++i){					double rad=(double)(rand()%10000+1)/10000.0*2.0*PI;					Point tp=p+T*Point(cos(rad),sin(rad));					if(tp.x>-EPS && tp.x-(double)X
         
          -EPS && tp.y-(double)Y
          
           bestnow){ bestnow=now; besttp=tp; } } } if(bestnow>EPS && (bestnow>ans || exp((bestnow-ans)/T)*10000.0>(double)(rand()%10000))){ ans=bestnow; p=besttp; } T*=0.9; } if(ans>allans){ allans=ans; allp=p; } } printf("The safest point is (%.1f, %.1f).\n",fabs(allp.x),fabs(allp.y)); } return 0;}