HDU 6097 数学几何

Mindis

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1519    Accepted Submission(s): 237
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

  

   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
  

 

Source

2017 Multi-University Training Contest - Team 6

 

题意 ： 给一个圆，两个点，这两个点不在圆外，两个点到圆心距离相等，求圆上点到两个点距离和的最小值。

思路 ： 首先要了解反演点

一般指二维反演中的点。

二维上反演以一个特定的反演圆为基础:圆心O为反演中心,圆半径为常数k,把点P反演为点P'就是使得OP×OP'=r^2 （来自百度百科）

首先做出两个点的反演点 P' 和 Q' ，ODP与ODP' 是相似三角型（这个不难证明）

所以极小的 PD + QD 可转化为极小的P'D + Q'D 

当P'Q'与圆有交点的时候可以使P'DQ'在同一直线上，这时P'Q'之间的距离就是 ODP' + ODQ'的最小值，那么ODP + ODQ的最小值就等于 P'Q' * （相似的比例）

当P'Q与圆无交点的时候，可以把P' 和 Q'当作一个椭圆的两个交点，我们知道椭圆上的点到椭圆上任意一点的距离之和都相等，所以我们慢慢放大椭圆直到与圆相切的时候，那个点就是最短距离点，你会发现，那个点一定在中垂线上，所以就去求中垂线与圆的交点的距离就可以了

#include<stdio.h>
#include<math.h>
#define eps 1e-8  

int main(){
	int T;
	double r,x1,y1,x2,y2,x3,y3,x4,y4;
	double d1,d2,d,k1,k2,dist,ans;
	double mx,my,smx,smy;
	scanf("%d",&T);
	while(T--){
		scanf("%lf %lf %lf %lf %lf",&r,&x1,&y1,&x2,&y2);
		
		d1 = sqrt(pow(x1,2) + pow(y1,2)); // 计算出其中一点到圆心距离 
		k1 = pow(r,2) / pow(d1,2);		// 这个 k 由 d(反演点) * d(p点) = r ^ 2 推出 
		x3 = x1 * k1,y3 = y1 * k1;		//用这个比例得出反演点坐标 
		x4 = x2 * k1,y4 = y2 * k1;
		
		mx = (x3 + x4) / 2;				//求出两反演点的中点 
		my = (y3 + y4) / 2;
		d2 = sqrt(pow(mx,2) + pow(my,2)); //中点到圆心的距离可以判断是否与圆相交 
		
		if(fabs(d1) <= eps){
			ans = 2 * r;
		}else if(d2 <= r){
			dist = sqrt(pow(x3 - x4,2) + pow(y3 - y4,2));
			k2 = d1 / r;
			ans = dist * k2;
		}else if(d2 > r){
			k2 = r / d2;
			smx = mx * k2,smy = my * k2;
			ans = 2 * (sqrt(pow(smx - x1,2) + pow(smy - y1,2)));
		}
		printf("%.7f\n",ans);
	}
	return 0;
}