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 exceed 10−6.
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
4
4
4 0
0 4
4
0 3
3 0
4
0 2
2 0
4
0 1
1 0
Sample Output
5.6568543
5.6568543
5.8945030
6.7359174
题意：有一个圆，圆心在原点，输入半径 r ，圆内有两个点P ， Q 到圆心距离相同，现在求在圆上找一点M，使得PM+QM最小？
思路：
对于圆内的两个点 P 和 Q ，如果以其作为椭圆两个焦点，由图一可以看到：当椭圆顶点与圆相切时 ， 圆与椭圆会有三个焦点，而椭圆上的点到P、Q的距离和是定值。那么可以缩小椭圆，接下来会有四个交点，继续缩小椭圆，得到图二，椭圆与圆只有两个交点，因为椭圆外的点到两个焦点的距离和大于2a，而椭圆上的点到两个焦点的距离和为2a,。在图2中，除了两个交点外，其余圆上点均在椭圆外，所以这时的交点到P、Q两点的距离和最小，等于2a。
如何找到这个相切的椭圆呢？二分解决。
具体：由P、Q两个焦点，我们可以确定c=PQ/2 , 现在如果再给出一个b值那么就能确定这个椭圆了，而b值越大椭圆越大，b值越小椭圆越小，所以我们要找到如图2所示，相切只有两个焦点时的b值，那么我们需要找的就是圆和椭圆相交时最小的b。我们可以求出图3中所示的 h 和 R -h ，那么椭圆的方程为：x^2/a^2 + (y-h)^2/b^2 = 1 ( 这个方程是PQ平行于X轴时的，但PQ不平行于X时也不影响，我们只是用它判断是否与圆相交 ) 圆：x^2 + y^2 =R^R , 并且b>0（显然），b<R-h ，所以在（0，R-h） 二分查找b。
代码如下：
#include <iostream>
#include <stdio.h>
#include <math.h>
#include <algorithm>
using namespace std;
const double eps = 1e-10;
double dis(double x1,double y1,double x2,double y2 )
{
return sqrt((x2-x1)*(x2-x1)+(y2-y1)*(y2-y1));
}
namespace IO {
const int MX = 4e7; //1e7占用内存11000kb
char buf[MX]; int c, sz;
void begin() {
c = 0;
sz = fread(buf, 1, MX, stdin);
}
inline bool read(int &t) {
while(c < sz && buf[c] != ‘-‘ && (buf[c] < ‘0‘ || buf[c] > ‘9‘)) c++;
if(c >= sz) return false;
bool flag = 0; if(buf[c] == ‘-‘) flag = 1, c++;
for(t = 0; c < sz && ‘0‘ <= buf[c] && buf[c] <= ‘9‘; c++) t = t * 10 + buf[c] - ‘0‘;
if(flag) t = -t;
return true;
}
}
int main()
{
IO::begin();
int T;
double R,x1,x2,y1,y2;
double x3,y3;
double A,B,C,dt,ans1,ans2;
//cin>>T;
IO::read(T);
while(T--)
{
//scanf("%lf%lf%lf%lf%lf",&R,&x1,&y1,&x2,&y2);
int xr, xx1, xx2, yy1, yy2;
IO::read(xr);
IO::read(xx1);
IO::read(yy1);
IO::read(xx2);
IO::read(yy2);
R = xr;
x1 = xx1;y1 = yy1;x2 = xx2;y2 = yy2;
double c=dis(x1,y1,x2,y2)*0.5;
c=c*c;
x3=(x1+x2)*0.5;
y3=(y1+y2)*0.5;
double h=dis(x3,y3,0.0,0.0);
//cout<<h<<endl;
double Rb=R-h;
double Lb=0.0,b,a,mid;
for(int i=0; i<30; i++)
{
mid=(Lb+Rb)*0.5;
b = mid;
b=b*b;
a=c+b;
A=a-b;
B=2.0*a*h;
C=b*R*R+a*h*h-a*b;
dt=B*B-4.0*A*C;
int flag=1;
if(dt>=-eps)
{
ans1=(-B+sqrt(B*B-4.0*A*C))*0.5/A;
ans2=(-B-sqrt(B*B-4.0*A*C))*0.5/A;
ans1=ans1*ans1;
ans2=ans2*ans2;
if(R*R-ans1>=-eps||R*R-ans2>=-eps)
flag=0;
}
if(flag==0)
Rb=mid;
else
Lb=mid;
}
printf("%.10f\n",sqrt(a)*2.0);
}
return 0;
}