HDU 6097 Mindis

Mindis

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1525    Accepted Submission(s): 239
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

444 00 440 33 040 22 040 11 0

 

Sample Output

5.65685435.65685435.89450306.7359174

 

Source

2017 Multi-University Training Contest - Team 6

题意：

在平面直角坐标系中，一个圆的圆心在原点

在该圆上或圆内给出两点P,Q，使得两点到圆心距离相等 PO = QO

现在，在圆上找出一点D，使得PD+QD的值最小

题解：

做P，Q反演点P'，Q'

连接P'Q'，做中垂线

（以P‘Q’为焦点做椭圆，存在椭圆的几何性质，椭圆上任一点到两焦点之和距离相等）

中垂线与圆的交点即为所求D点

再根据相似三角形的几何性质，求出比例关系 对应相乘即可

（思路参考kkkkahlua 谢谢哥）

#include<stdio.h>#include<string.h>#include<algorithm>#include<math.h>#define ll long longusing namespace std;const double eps = 1e-8;int main(){int t;scanf("%d",&t);while(t--){double r,x1,y1,x2,y2,ans;scanf("%lf %lf %lf %lf %lf",&r,&x1,&y1,&x2,&y2);double d0 = sqrt(x1 * x1 + y1 * y1);if(fabs(d0) < eps){printf("%.7lf\n",2 * r);continue;}//利用相似三角形比例关系证明 double k = r * r / (d0 * d0);double px = x1 * k,py = y1 * k;double qx = x2 * k,qy = y2 * k;double midx = (px + qx) / 2;double midy = (py + qy) / 2;double d = sqrt(midx * midx + midy * midy);if(d <= r){//与圆有交点 利用P’Q’与O的距离和半径进行比较 double distance = sqrt(pow(px - qx,2) + pow(py - qy,2));ans = distance * d0 / r;}else{double sumx = midx * (r / d);double sumy = midy * (r / d);ans = 2 * sqrt(pow(sumx - x1,2) + pow(sumy - y1,2));}printf("%.7lf\n",ans);}return 0;}