POJ3384+半平面交

/*
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 ){//>=0
			tp[ ++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;
}