hdu 4454 Stealing a Cake

简单的计算几何；

可以把0-2*pi分成几千份，然后找出最小的；

也可以用三分；

 #include<cstdio>
 #include<cmath>
 #include<algorithm>
 #define pi acos(-1)
 #define eps 1e-6
 using namespace std;

 struct node
 {
     double x,y;
     node(double x=,double y=):x(x),y(y){ }
     bool operator<(const node &t)const
     {
         if(x==t.x)return y<t.y;
         else return x<t.x;
     }
 };
 node operator-(node u,node v){return node(u.x-v.x,u.y-v.y);}
 node operator+(node u,node v){return node(u.x+v.x,u.y+v.y);}
 node operator*(node u,double v){return node(u.x*v,u.y*v);}
 node operator/(node u,double v){return node(u.x/v,u.y/v);}
 int dcmp(double x)
 {
     if(fabs(x)<eps)return ;
     else return x<?-:;
 }
 bool operator==(const node& a,const node& b)
 {
     return dcmp(a.x-b.x)== && dcmp(a.y-b.y)==;
 }
 double dot(node a,node b){return a.x*b.x+a.y*b.y;}
 double length(node a){return sqrt(dot(a,a));}
 double cross(node a,node b){return a.x*b.y-a.y*b.x;}
 double disofnode(node p,node a,node b)
 {
     if(a==b)return length(p-a);
     node v1=b-a,v2=p-a,v3=p-b;
     if(dcmp(dot(v1,v2))<)return length(v2);
     else if(dcmp(dot(v1,v3))>)return length(v3);
     else return fabs(cross(v1,v2))/length(v1);
 }

 node a,yuan,d[];
 double r;
 double dis(double alph)
 {
     node tmp;
     tmp.x=yuan.x+r*cos(alph);
     tmp.y=yuan.y+r*sin(alph);
     double dd1=min(disofnode(tmp,d[],d[]),disofnode(tmp,d[],d[]));
     double dd2=min(disofnode(tmp,d[],d[]),disofnode(tmp,d[],d[]));
     double ret=min(dd1,dd2);
     return (ret+length(a-tmp));
 }

 int main()
 {
     while(scanf("%lf%lf",&a.x,&a.y))
     {
         if(dcmp(a.x)== && dcmp(a.y)==)break;
         scanf("%lf%lf",&yuan.x,&yuan.y);
         scanf("%lf",&r);
         double x1,x2,y1,y2;
         scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
         d[].x=x1,d[].y=y1;
         d[].x=x2,d[].y=y1;
         d[].x=x2,d[].y=y2;
         d[].x=x1,d[].y=y2;
         sort(d,d+);
         double mi=0.0,ma=*pi;
         double ans=9999999999.0;
         for(int i=;i<=;i++)
         {
             double mm=(*pi*i)/;
             double fd=dis(mm);
             ans=min(ans,fd);
 //            double m1=mi+(ma-mi)/3;//三分的代码
 //            double m2=ma-(ma-mi)/3;
 //            if(dis(m1)>dis(m2))mi=m1;
 //            else ma=m2;
         }
         printf("%.2lf\n",ans);
     }
     return ;
 }