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<cstdio> #include<cmath> #include<algorithm> #include<cstdlib> 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>-EPS && tp.y-(double)Y<EPS){ double now=calc(tp); if(now>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; }
时间: 2024-10-12 20:05:00