hdu 6097 Mindis(几何反演)

Mindis

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2598    Accepted Submission(s): 497
Special Judge

Problem Description

The center coordinate of the circle C is O, the coordinate of O is (0,0) , and the radius is r.
P and Q are two points not outside the circle, and PO = QO.
You need to find a point D on the circle, which makes PD+QD minimum.
Output minimum distance sum.

 

Input

The first line of the input gives the number of test cases T; T test cases follow.
Each case begins with one line with r : the radius of the circle C.
Next two line each line contains two integers x , y denotes the coordinate of P and Q.

Limits
T≤500000
−100≤x,y≤100
1≤r≤100

 

Output

For each case output one line denotes the answer.
The answer will be checked correct if its absolute or relative error doesn't exceed10−6.
Formally, let your answer be a, and the jury's answer be b. Your answer is considered correct if|a−b|max(1,b)≤10−6.

 

Sample Input

444 00 440 33 040 22 040 11 0

 

Sample Output

5.6568543
5.6568543
5.8945030
6.7359174

根据几何反演，将OP,OQ沿P,Q延长，使OP*OP'=r*r,OQ*OQ'=r*r

△ODP∽△OP'D  x=DP/DP'=OP/OD=OD/OP'=|OP|/r

P'Q'/PQ=OP'/OP=(r*r)/(|OP|*|OP|)

(P'D+Q'D)=(PD*x+QD*x)=x*(ans)

ans为所求答案

第一种情况：P'Q'与圆有交点
D是P'Q'与圆的交点

P'D'+Q'D'=P'Q'

第二种情况：P'Q'与圆没有交点
D是P'Q'中垂线与圆的交点（因为P,Q是关于O中心对称的，所以P',Q'也与0中心对称，所以P'Q'的中垂线一定过O）
W是P'Q'的中点

P'D'+Q'D=2*P'D=2*√P'Q'^2+(OW-r)^2

#include<cstdio>#include<cstring>#include<stdlib.h>#include<cmath>using namespace std;int r,px,py,qx,qy;int main(){int t;scanf("%d",&t);while(t--){scanf("%d",&r);scanf("%d%d",&px,&py);scanf("%d%d",&qx,&qy);long long int a,b;a=px*px+py*py;    //OP,OQb=(px-qx)*(px-qx)+(py-qy)*(py-qy);  //PQdouble ans;if(r*r*(4*a-b)<=4*a*a)    //p'q'与圆有交点{if(a==0) ans=(double)2*r;   //若p,q都在圆心else{double pq=r*r*b*1.0/a;   ans=sqrt(pq);}}else{double tmp=r*r+a-r*sqrt(4*a-b);ans=sqrt(tmp)*2;}printf("%.7lf\n",ans);}return 0;}