poj1375---計算幾何

//分別計算每個圓在floor上的陰影部分，然後貪心一下就好了
//在求切線時，是把圓點逆時針旋轉angle和2*pi-angle度

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
#include<queue>
#define dist(a,b) sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y))
#define cross(a,b,c) (b.x-a.x)*(c.y-a.y)-(b.y-a.y)*(c.x-a.x)
#define dot(a,b,c) (b.x-a.x)*(c.x-a.x)+(b.y-a.y)*(c.y-a.y)
#define delt(a) fabs(a)<eps?0:a>0?1:-1
#define pi acos(-1.0)
#define eps 1e-8
#define inf 1e20
#define N 1005
using namespace std;
int n,m,t,nl,ml;
struct TPoint
{
	double x,y,r;
}pt[N],st,ans[N];
TPoint ps,pe;
bool scan()
{
	scanf("%d",&n);
	if(n==0) return 0;
	scanf("%lf%lf",&st.x,&st.y);
	for(int i=0;i<n;i++) scanf("%lf%lf%lf",&pt[i].x,&pt[i].y,&pt[i].r);
	return 1;
}
double segcross(TPoint a,TPoint b,TPoint c,TPoint d)
{
	double s1=cross(a,c,d),s2=cross(b,c,d);
	return (s2*a.x-s1*b.x)/(s2-s1);
}
void solve(int id)
{
	double ang=asin(pt[id].r/dist(st,pt[id]));
	double ang1=2*pi-ang;
	TPoint p1,p2,p3;
	p1.x=st.x+(pt[id].x-st.x)*cos(ang1)-(pt[id].y-st.y)*sin(ang1);
	p1.y=st.y+(pt[id].x-st.x)*sin(ang1)+(pt[id].y-st.y)*cos(ang1);
	p2.x=st.x+(pt[id].x-st.x)*cos(ang)-(pt[id].y-st.y)*sin(ang);
	p2.y=st.y+(pt[id].x-st.x)*sin(ang)+(pt[id].y-st.y)*cos(ang);
	ans[id].x=segcross(st,p1,ps,pe);
	ans[id].y=segcross(st,p2,ps,pe);
}
bool cmp(TPoint a,TPoint b)
{
	return a.x<b.x||(a.x==b.x&&a.y<b.y);
}
void solve()
{
	int i,j,k;
	ps.x=ps.y=pe.y=0,pe.x=2;
	for(i=0;i<n;i++) solve(i);
	sort(ans,ans+n,cmp);
	double ml=ans[0].x,mr=ans[0].y;
	for(i=1;i<n;i++)
	{
		if(mr+eps<ans[i].x)
		{
			printf("%.2f %.2f\n",ml,mr);
			ml=ans[i].x,mr=ans[i].y;
		}
		else mr=max(mr,ans[i].y);
	}
	printf("%.2f %.2f\n",ml,mr);
	puts("");
}
int main()
{
	while(scan())
	{
		solve();
	}
	return 0;
}