poj 1375 Intervals 圆的切线

题目链接：http://poj.org/problem?id=1375

题意：有一光源，被若干个圆摭住了，求地面上阴影

题解：利用向量旋转求出切线与圆的两个交点，根据两点成线得出aX+bY+c=0的直线，令Y=0时，求出地面的坐标，再合并有连接的阴影。

//代码如下：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<math.h>
#define eps 1e-8

//using namespace std;
const int maxn=1000;
struct point{double x,y;};
struct circle {point cen;double r;}cir[maxn];
struct Node{double s,e;}node[maxn];

double f[maxn][2];
point st;
double distance(point p1,point p2)
{
	return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y));
}
void getline(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;
}

void cirinser(circle c,point &inser1,point &inser2)//圆与切线交点(保证st在圆外)

{
	point dir;
	double dis=distance(c.cen,st);
	double temp=sqrt(dis*dis-c.r*c.r);
	double sina=temp/dis;
	double cosa=c.r/dis;
	dir.x=(st.x-c.cen.x)/dis*c.r;
	dir.y=(st.y-c.cen.y)/dis*c.r;
	inser1.x=c.cen.x+(dir.x*cosa-dir.y*sina);
	inser1.y=c.cen.y+(dir.x*sina+dir.y*cosa);
	
	inser2.x=c.cen.x+(dir.x*cosa+dir.y*sina);
	inser2.y=c.cen.y+(-dir.x*sina+dir.y*cosa);
}
bool zero(double x){return x>0? x<eps:x>-eps;}
bool cmp( Node a, Node b)
{
	if(zero(a.s-b.s))return a.e-b.e<0;
	else return a.s-b.s<0;
}
int main()
{
	int n;
	while(scanf("%d",&n),n)
	{
		scanf("%lf%lf",&st.x,&st.y);
		int i;
		point inser1,inser2;
		double a,b,c;
		for(i=0;i<n;i++)
		{
			scanf("%lf%lf%lf",&cir[i].cen.x,&cir[i].cen.y,&cir[i].r);
			cirinser(cir[i],inser1,inser2);
			getline(st,inser1,a,b,c);
			node[i].s=-c/a;
			getline(st,inser2,a,b,c);
			node[i].e=-c/a;
			if(node[i].s-node[i].e>0)std::swap(node[i].s,node[i].e);
		}
		std::sort(node,node+n,cmp);
		
		f[0][0]=node[0].s;
		f[0][1]=node[0].e;
		int cnt=0;
		for(i=1;i<n;i++)
		{
			if(node[i].s-f[cnt][1]<eps &&node[i].e-f[cnt][1]>0)f[cnt][1]=node[i].e;
			else if(node[i].s-f[cnt][1]>0)
			{
				f[++cnt][0]=node[i].s;
				f[cnt][1]=node[i].e;
			}
		}
		for(i=0;i<=cnt;i++)
			printf("%.2lf %.2lf\n",f[i][0],f[i][1]);
		printf("\n");
	}
	return 0;
}