HDU6097-Mindis

Mindis

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2678 Accepted Submission(s): 524
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

题目大意：给出圆的半径r和圆内等距两点P、Q，在圆上找一点D，问P、Q到D的距离之和最小是多少？
解题思路：作P、Q的反演点，即 OP∗OP′=r2，OQ∗OQ′=r2 ，根据 P′Q′ 与圆是否相交来判断D的位置，若相交则结果为 P′Q′ 乘上比例 OP/r ，若不相交，则先求出 P′Q′ 的中点 D′ ，在根据比例确定 D 的位置，计算 PD+QD 即可。

#include
#include
#include
using namespace std;
const double eps=1e-8;
struct node
{
  double x,y;
}p[10];

int dcmp(double x)
{
  if(fabs(x)<=eps) return 0;
  return x<0?-1:1;
}

double calc(double x1,double y1,double x2,double y2)
{
  return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
}

int main()
{
    int T;
    scanf("%d",&T );
    while (T--) {
      double r;
      scanf("%lf",&r );
      scanf("%lf%lf",&p[1].x,&p[1].y );
      scanf("%lf%lf", &p[2].x,&p[2].y);
      double op=calc(p[1].x,p[1].y,0,0);
      if(dcmp(op)==0)
      {printf("%.7f\n",2*r );continue;}
      double k=r*r/op/op;
      p[3].x=p[1].x*k;p[3].y=p[1].y*k;
      p[4].x=p[2].x*k;p[4].y=p[2].y*k;
      p[5].x=(p[3].x+p[4].x)/2;
      p[5].y=(p[3].y+p[4].y)/2;
      double od=calc(p[5].x,p[5].y,0,0);
      if(od<=r)
      {
        printf("%.7f\n",calc(p[3].x,p[3].y,p[4].x,p[4].y)*op/r );
      }else
      {
        p[6].x=p[5].x*r/od;p[6].y=p[5].y*r/od;
        printf("%.7f\n",2*calc(p[6].x,p[6].y,p[1].x,p[1].y));
      }
    }
    return 0;
}