/*给出一个凸多边形的房间，根据风水要求，把两个圆形地毯铺在房间里，不能折叠，不能切割，可以重叠。问最多能覆盖多大空间，输出两个地毯的圆心坐标。多组解输出其中一个将多边形的边内移R之后，半平面交区域便是可以放入圆的可行区域*/#include<stdio.h>#include<string.h>#include<stdlib.h>#include<math.h>#include<algorithm>using namespace std;const int maxn = 105;const int maxm = 1005;const double eps = 1e-5;const double pi = acos(-1.0);struct Point{double x,y;};struct Line{Point a,b;};Point pnt[ maxn ],res[ maxm ],tp[ maxm ];double xmult( Point op,Point sp,Point ep ){return (sp.x-op.x)*(ep.y-op.y)-(sp.y-op.y)*(ep.x-op.x);}double dist( Point a,Point b ){return sqrt( (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y) );}void Get_equation( Point p1,Point p2,double &a,double &b,double &c ){a = p2.y-p1.y;b = p1.x-p2.x;c = p2.x*p1.y-p1.x*p2.y;}//直线方程Point Intersection( Point p1,Point p2,double a,double b,double c ){double u = fabs( a*p1.x+b*p1.y+c );double v = fabs( a*p2.x+b*p2.y+c );Point tt;tt.x = (p1.x*v+p2.x*u)/(u+v);tt.y = (p1.y*v+p2.y*u)/(u+v);return tt;}//交点、按照三角比例求出交点double GetArea( Point p[],int n ){double sum = 0;for( int i=2;i<n;i++ ){sum += xmult( p[1],p[i],p[i+1] );}return -sum/2.0;}//面积，顺时针为正void cut( double a,double b,double c,int &cnt ){int temp = 0;for( int i=1;i<=cnt;i++ ){if( a*res[i].x+b*res[i].y+c>-eps ){//>=0tp[ ++temp ] = res[i];}else{if( a*res[i-1].x+b*res[i-1].y+c>eps ){tp[ ++temp ] = Intersection( res[i-1],res[i],a,b,c );}if( a*res[i+1].x+b*res[i+1].y+c>eps ){tp[ ++temp ] = Intersection( res[i],res[i+1],a,b,c );}}}for( int i=1;i<=temp;i++ )res[i] = tp[i];res[ 0 ] = res[ temp ];res[ temp+1 ] = res[ 1 ];cnt = temp;}void solve( int n,double r ){if( GetArea( pnt,n)<eps )reverse( pnt+1,pnt+1+n );pnt[0] = pnt[n];pnt[n+1] = pnt[1];for( int i=0;i<=n+1;i++ )res[ i ] = pnt[ i ];int cnt = n;for(int i=1;i<=n;i++){ double a,b,c; Point p1,p2,p3; p1.y=pnt[i].x-pnt[i+1].x;p1.x=pnt[i+1].y-pnt[i].y; double k=r/sqrt(p1.x*p1.x+p1.y*p1.y); p1.x=k*p1.x;p1.y=p1.y*k; p2.x=p1.x+pnt[i].x;p2.y=p1.y+pnt[i].y; p3.x=p1.x+pnt[i+1].x;p3.y=p1.y+pnt[i+1].y; //移动R的部分 Get_equation( p2,p3,a,b,c ); cut(a,b,c,cnt); } double max_dis = 0;Point s,t;for( int i=1;i<=cnt;i++ ){for( int j=1;j<=cnt;j++ ){double d = dist( res[i],res[j] );if(d+eps>max_dis ){max_dis = d;s = res[i];t = res[j];}}}printf("%.4lf %.4lf %.4lf %.4lf\n",s.x,s.y,t.x,t.y);}int main(){int n;double r;while( scanf("%d%lf",&n,&r)==2 ){for( int i=1;i<=n;i++ ){scanf("%lf%lf",&pnt[i].x,&pnt[i].y);}solve( n,r );}return 0;}
