Mindis（hdu6097）

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

5.65685435.65685435.8945030
6.7359174
给一个圆C和圆心O，P、Q是圆上或圆内到圆心距离相等的两个点，在圆上取一点D，求|PD| + |QD|的最小值
做点P、Q的反演点P'、Q', |OP| * |OP'| = r ^ 2, |OQ| * |OQ'| = r ^ 2.
|OP| / r = r / |OP'|, 即|OP| / |OD| = |OD| / |OP'|, 所以△OPD ∽ △ODP', 同理, △OQD ∽ △ODQ', 相似比为|OP| / r.
所以, |PD| + |QD| = (|P'D| + |Q'D|) * |OP| / r. 转化为求|P'D| + |Q'D|的最小值
若P'Q'与圆C有交点,则|P'D| + |Q'D|的最小值为|P'Q'| (此时D为P'Q'与圆C的交点, |P'D| + |Q'D| 为直线,值最小)
若P'Q'与圆C无交点,则D为PQ的中垂线与圆C的交点时, |P'D| + |Q'D|取得最小值(以P、Q为焦点做椭圆，令椭圆不断变大，椭圆与圆C有一个交点时,此交点即为点D)
这道题要用比例做 ，要用斜率找点很麻烦，有斜率等于0 或者没有斜率的情况
#include <iostream>#include <cstdio>#include <cmath>using namespace std;const double eps=1e-8;struct point{    double x,y;}p,q;double len(point a,point c){    return sqrt((a.x-c.x)*(a.x-c.x)+(a.y-c.y)*(a.y-c.y));}int main(){    int T;    double r;    scanf("%d", &T);    while (T-- )    {        scanf("%lf",&r);        scanf("%lf%lf",&p.x,&p.y);        scanf("%lf%lf",&q.x,&q.y);        point o;        o.x=0.0;        o.y=0.0;        double op=len(p,o);        point pp,qq,d;        if(abs(op)<=eps)        {            printf("%.6lf\n",r*2);        }        else        {            double l=r*r/op;            pp.x=p.x*l/op;            pp.y=p.y*l/op;            qq.x=q.x*l/op;            qq.y=q.y*l/op;            d.x=(pp.x+qq.x)/2;            d.y=(pp.y+qq.y)/2;            if(len(o,d)<=r)            {                double ans=len(pp,qq)*op/r;                printf("%.6lf\n",ans);            }            else            {                double dis=len(o,d);                double t=r/dis;                d.x=d.x*t;                d.y=d.y*t;                double ans=2*len(d,p);                printf("%.6lf\n",ans);            }        }    }    return 0;}