【计算几何】【预处理】【枚举】Urozero Autumn Training Camp 2016 Day 5: NWERC-2016 Problem K. Kiwi Trees

发现由于角的度数和边的长度有限制，那俩圆如果放得下的话，必然是塞在两个角里。

于是预处理n个圆心的位置（注意要判断那个圆会不会和其他的边界相交），然后n^2枚举俩角即可。

#include<cstdio>
#include<cmath>
using namespace std;
#define EPS 0.00000001
struct Point{
	double x,y;
	double length(){
		return sqrt(x*x+y*y);
	}
}a[2010];
typedef Point Vector;
Vector unit(Vector v){
	double l=v.length();
	return (Vector){v.x/l,v.y/l};
}
Vector operator - (const Point &a,const Point &b){
	return (Vector){a.x-b.x,a.y-b.y};
}
Vector operator + (const Point &a,const Point &b){
	return (Vector){a.x+b.x,a.y+b.y};
}
double dot(const Vector &a,const Vector &b){
	return a.x*b.x+a.y*b.y;
}
double Cross(const Vector &a,const Vector &b){
	return a.x*b.y-a.y*b.x;
}
Vector operator * (const double &K,const Vector &v){
	return (Vector){K*v.x,K*v.y};
}
int n;
Point calc(int I){
	double jiao=acos(dot(a[I+1]-a[I],a[I-1]-a[I])/(a[I+1]-a[I]).length()/(a[I-1]-a[I]).length());
	double d=4000.0/tan(jiao/2.0);
	double l=4000.0/sin(jiao/2.0);
	Point p1=a[I]+d*unit(a[I+1]-a[I]);
	Point p2=a[I]+d*unit(a[I-1]-a[I]);
	Point M=(Point){(p1.x+p2.x)/2.0,(p1.y+p2.y)/2.0};
	return a[I]+l*unit(M-a[I]);
}
double DisToSegment(Point P,Point A,Point B)
{
	Vector v1=B-A,v2=P-A,v3=P-B;
	if(dot(v1,v2)<EPS) return v2.length();
	else if(dot(v1,v3)>EPS) return v3.length();
	else return fabs(Cross(v1,v2))/v1.length();
}
Point yuanxins[2010];
bool can[2010];
int main(){
//	freopen("k.in","r",stdin);
	scanf("%d",&n);
	for(int i=1;i<=n;++i){
		scanf("%lf%lf",&a[i].x,&a[i].y);
	}
	a[++n]=a[1];
	for(int i=n;i>=1;--i){
		a[i+1]=a[i];
	}
	a[1]=a[n];
	++n;
	for(int i=2;i<n;++i){
		yuanxins[i]=calc(i);
		can[i]=1;
		for(int j=2;j<n;++j){
//			double tmp=DisToSegment(yuanxins[i],a[j],a[j+1]);
			if(DisToSegment(yuanxins[i],a[j],a[j+1])-4000.0<-EPS){
				can[i]=0;
				break;
			}
		}
	}
	for(int i=2;i<n;++i){
		if(Cross(a[i+1]-a[i],a[i]-a[i-1])>EPS && can[i]){
			Point A=yuanxins[i];
			for(int j=i+1;j<n;++j){
				if(Cross(a[j+1]-a[j],a[j]-a[j-1])>EPS && can[j]){
					Point B=yuanxins[j];
//					double tmp=(B-A).length();
					if((B-A).length()-8000.0>-EPS){
						printf("%.8f %.8f\n%.8f %.8f\n",A.x,A.y,B.x,B.y);
						return 0;
					}
				}
			}
		}
	}
	puts("impossible");
	return 0;
}