HDU 6097Mindis(利用椭圆二分)

Mindis

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 1887    Accepted Submission(s): 321
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 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

<
d11f
/pre>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：在圆内有两点P,Q，圆心为O，PO=QO，给你圆的半径r和P,Q的坐标，求圆上一点到P,Q，两点距离最小值
思路：若以P,Q为焦点做一椭圆若椭圆与圆相交，那么交点处到P,Q的距离即为椭圆方程的2a，我们利用椭圆上任意一点到焦点的距离都为2a这一性质去找最小2a就为答案。
圆与椭圆相交的极限情况为椭圆与圆内切，此时的2a就是答案
对于所给的P,Q两点我们可以旋转坐标轴使得PQ直线与x轴平行，因为圆的圆心固定所以旋转并不会影响答案。
然后我们以P,Q为焦点做椭圆方程 

，h为PQ直线到原点距离


联立两个方程化简得


设



当

时方程有解但是这个解可能为增根
利用求根公式求得


,


因为y1,y2的有效范围为[-r,r]所以还要判断一下有几个符合的实根
然后我们只用用二分来枚举b，然后判断有几个根就行
一：

二：


三：

四：


当y有两个不同的根时实际上b是大了所以符合的b其实在l到mid这一段区间(图一)，当y没有根时b应该在mid到r这段区间内(图二)，当y为有一个根时跳出即可(图三)，而二分的右边界r应该是椭圆与圆的上顶点相切这种情况(图四)，明显的此时b=r-h
故二分的范围为0~r-h
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
const double eps=1e-10;
double r,c,h;
struct point
{
double x,y;
}p,q;
double dis(point a,point b)
{
return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
int judge(double b)
{
double a=sqrt(b*b+c*c);
double A=a*a-b*b;
double B=-2*h*a*a;
double C=a*a*h*h+b*b*r*r-a*a*b*b;
double delta=B*B-4*A*C;
int num=0;
if(delta>=0)
{
double y1=(-B+sqrt(delta))/(2*A);
double y2=(-B-sqrt(delta))/(2*A);
if(y1<=r&&y1>=-r)
num++;
if(y2<=r&&y2>=-r)
num++;
}
return num;
}
double binarysearch(double l,double r)
{
double mid;
while(l+eps<r)
{
mid=(l+r)/2;
int flag=judge(mid);
if(flag>1)
r=mid;
else
l=mid;
if(flag==1)
break;
}
return mid;
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
scanf("%lf%lf%lf%lf%lf",&r,&p.x,&p.y,&q.x,&q.y);
c=dis(p,q)/2;
h=sqrt(p.x*p.x+p.y*p.y-c*c);
double b=binarysearch(0,r-h);
double a=sqrt(b*b+c*c);
printf("%.7lf\n",2*a);
}
return 0;
}