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；

1 #include
      
        2 #include
       
         3 #include
        
          4 #define pi acos(-1) 5 #define eps 1e-6 6 using namespace std; 7  8 struct node 9 {10     double x,y;11     node(double x=0,double y=0):x(x),y(y){ }12     bool operator<(const node &t)const13     {14         if(x==t.x)return y
         
          0)return length(v3);40     else return fabs(cross(v1,v2))/length(v1);41 }42 43 44 45 node a,yuan,d[4];46 double r;47 double dis(double alph)48 {49     node tmp;50     tmp.x=yuan.x+r*cos(alph);51     tmp.y=yuan.y+r*sin(alph);52     double dd1=min(disofnode(tmp,d[0],d[1]),disofnode(tmp,d[0],d[2]));53     double dd2=min(disofnode(tmp,d[1],d[3]),disofnode(tmp,d[2],d[3]));54     double ret=min(dd1,dd2);55     return (ret+length(a-tmp));56 }57 58 int main()59 {60     while(scanf("%lf%lf",&a.x,&a.y))61     {62         if(dcmp(a.x)==0 && dcmp(a.y)==0)break;63         scanf("%lf%lf",&yuan.x,&yuan.y);64         scanf("%lf",&r);65         double x1,x2,y1,y2;66         scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);67         d[0].x=x1,d[0].y=y1;68         d[1].x=x2,d[1].y=y1;69         d[2].x=x2,d[2].y=y2;70         d[3].x=x1,d[3].y=y2;71         sort(d,d+4);72         double mi=0.0,ma=2*pi;73         double ans=9999999999.0;74         for(int i=0;i<=5000;i++)75         {76             double mm=(2*pi*i)/5000;77             double fd=dis(mm);78             ans=min(ans,fd);79 //            double m1=mi+(ma-mi)/3;//三分的代码80 //            double m2=ma-(ma-mi)/3;81 //            if(dis(m1)>dis(m2))mi=m1;82 //            else ma=m2;83         }84         printf("%.2lf\n",ans);85     }86     return 0;87 }