HDU 6097 Mindis 几何

题目链接：HDU 6097

Mindis

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 3168    Accepted Submission(s): 644
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

 

题意：

一个圆和2个点PQ，PQ在园内并且OP=OQ，然后找圆上一点D使得PD+QD最小。

题目分析：

找PQ的反演点然后巧妙的构造相似三角形，很巧妙的解决了精度和时间问题。给“优秀的黄金分割”卡过的dalao们递茶～

////  main.cpp//  HDU6097 Mindis////  Created by teddywang on 2017/8/13.//  Copyright © 2017年 teddywang. All rights reserved.//#include <iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>using namespace std;double xp,yp,xq,yq,r,R;double x,y,x2,y2;double eps=1e-8;int T;struct point{    double x,y;    point(){        x=0;        y=0;    }    point(double X,double Y){        x=X;        y=Y;    }    };point p,q;point zero=point(0.0,0.0);double dis(point a,point b){    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));}point shucheng(point a,double b){    point c=point(a.x*b,a.y*b);    return c;}double chacheng(point a,point b){    double buf=(a.x*b.y)-(a.y*b.x);    if(buf>eps) return buf;    else return 0-buf;}bool judge_eq(point a,point b){    if(a.x==b.x&&a.y==b.y) return true;    else return false;}bool judge2(point a,point b){    double buf=r*r/dis(zero,a)/dis(zero,b);    point c=shucheng(a,buf);    point d=shucheng(b,buf);    double s=chacheng(c,d)/dis(c,d);    if(s-r>eps) return false;    else return true;}int main(){    scanf("%d",&T);    double ans;    while(T--)    {        scanf("%lf",&r);        scanf("%lf%lf",&p.x,&p.y);        scanf("%lf%lf",&q.x,&q.y);        if(judge_eq(p, q))        {            ans=(r-dis(zero,p))*2;        }        else        {            if(judge2(p,q))            {                double buf=r*r/dis(zero,p)/dis(zero,p);                point p1=shucheng(p,buf);                point q1=shucheng(q,buf);                ans=dis(p1,q1)*dis(p,zero)/r;            }            else            {                point mid=point((p.x+q.x)/2.00,(p.y+q.y)/2.00);                double buf=r/dis(zero,mid);                mid=shucheng(mid,buf);                ans=dis(mid,p)+dis(mid,q);            }        }        printf("%.7f\n",ans);    }}