圆的反演

n+e posted @ 2015年7月11日 18:35 in Interesting with tags 计算几何 数学 , 7402 阅读

题目大概是这样的:

有很多圆,满足 $C_n(n\ge 4)$ 与 $C_1,C_2,\cdots ,C_{n-1}$ 都相切,$C_3$ 与 $C_1,C_2$ 相切,$C_2$ 与 $C_1$ 相切,图形如下:

现已知 $r_1,r_2,r_3,r_4$,求 $r_n$。询问数<=1000w,$n\le 10^9$,时限 $1.5s$

例如:当 $r_1=6,r_2=r_3=3,r_4=2$ 的时候,圆大概长这样:

 

圆的反演是什么呢?

我们先选定一个点 $O$为反演中心,以 $O$ 为圆心,半径为 $r$ 画一个圆。然后对于平面上的点 $P$ 和 $P'$,如果 $P$ 和 $P'$ 在以 $O$ 为起点的射线上,并且 $|OP|\cdot|OP'|=r^2$,那么就说 $P$ 和 $P'$ 互为反演点。

所以圆外的点反演一下会到圆内,圆内会到圆外,圆上则不变。

我找了 $GeoGebra$ 来玩玩,效果差不多是长下面这个样子的:

反演中心为坐标原点,点 $H$ 在直线上移动,点 $J$ 为射线 $AH$的焦点,点 $I$ 为点 $H$ 的反演点。

(图片是可以点开来看的...)

可以看到,一条不过反演中心的直线反演之后变成了一个过反演中心的圆。而且直线和两个圆两两相交。

由于反演的可逆性,这个圆反演一下就变成了一条直线啦~

一个不过反演中心的圆反演以后是一个以反演中心位似的圆。

从上面的图来看,显然两个圆是反过来对应的,并不是直接位似。还有切记,圆心反演完以后并不是反演完的圆的圆心,比如上面的点 $H$ 和 $I$。

再来一个好玩点的:

抛物线反演一下,不知道变成了什么。。。右边的黑线是 $2\sqrt{x}$,然而不能拟合。

然后稍微往下移一点就变成了心型线!

椭圆好像跟上面两种差不多,不是太好玩。

双曲线反演一下变成了 $\infty$...

 

我们不妨把样例反演一下变成下面这样:

由于反演前后的几何性质是不会发生改变的,比如反演前两个圆相切,反演完以后他们两个还是相切,只不过可能不是圆与圆相切而是直线与圆相切之类的,因此我们可以像这张图这样建立反演中心,圆 $C_1,C_2$ 反演完以后就变成了两条直线,而 $C_3,C_4,\cdots$ 反演完之后,由于没过反演中心,所以还是圆;又因为它们都与圆 $C_1,C_2$ 相切,所以反演以后的圆统统都被夹在了两条直线里面,大小都一样,而且一个挨着一个。

这里详细讲了怎么求一个圆的反演

然而我觉得如果不会求的话,可以随便找圆上的三个点,然后把这三个点反一下,然后再以这三个点画个圆就好了,简单粗暴= =

注意不能直接求圆心。原因在上面第二张截屏。

于是这题的解法已经十分明了了:先把 $C_1,C_2$ 反成直线,解三角形解出 $C_3$ 的位置,然后反成小圆,看一下 $C_4$ 塞哪里符合题意,然后就可以 $O(1)$ 求出第 $n$ 个圆反演完以后的圆,直接反回去就好了。

#include<cstdio>
#include<cmath>
typedef double ld;
struct P{
	ld x,y;
	P operator*(ld a)const{return(P){x*a,y*a};}
	P operator/(ld a)const{return(P){x/a,y/a};}
}o3;
ld len2(const P&a){return a.x*a.x+a.y*a.y;}
P xorr(const P&a){return a/len2(a);}
int n,p,m,b,t;ld r,A,B,C,cosB,sinB,st,ans;
struct Cir{P o;ld r;}c[5],d[5];
Cir xorr(Cir a){
	ld d=sqrt(len2(a.o)),r=(1/(d-a.r)-1/(d+a.r))/2;
	a.o=a.o*(1/(d-a.r)-r)/d,a.r=r;
	return a;
}
int main(){
	scanf("%d%d%d%d%d",&t,&n,&p,&m,&b);
	scanf("%lf%lf%lf%lf",&c[1].r,&c[2].r,&c[3].r,&st);
	c[1].o.y=c[1].r;c[2].o.y=c[2].r;
	d[4].r=r=(xorr((P){c[2].r,c[2].r}).y-xorr((P){c[1].r,c[1].r}).y)*.5;
	A=c[1].r-c[2].r,B=c[2].r+c[3].r,C=c[1].r-c[3].r;
	cosB=(A*A+C*C-B*B)/2/A/C;sinB=sqrt(1-cosB*cosB);
	c[3].o.x=C*sinB;c[3].o.y=c[1].r-C*cosB;d[3]=xorr(c[3]);
	d[4].o=(P){d[3].o.x+r+r,d[3].o.y},c[4]=xorr(d[4]);
	if(std::abs(st-c[4].r)>1e-9)r=-r;
	for(r*=2;t--;){
		n=1LL*p*n%m+b;
		if(n<=4)ans+=c[n].r;else{
			d[4].o=(P){d[3].o.x+r*(n-3),d[3].o.y};
			register ld _d=sqrt(len2(d[4].o)),_r=(1/(_d-d[4].r)-1/(_d+d[4].r))/2;ans+=_r;
		}
	}
	printf("%lf\n",ans);
}
---By n+e
zzzz 说:
2017年8月11日 13:08

请问聚聚有这道题的地址吗,

Avatar_small
n+e 说:
2017年9月11日 21:55

@zzzz: CC2015 NTHCIR

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