Hdu6097 Mindis（2017多校第6场）

Mindis

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1056    Accepted Submission(s): 112
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

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题目的意思是给出一个圆，和圆内（上）距圆心的相等的两个点，求在圆上取一个点使得到这两个点的距离和最小

思路：数学推导出公式计算，注意卡精度（队友真是太强了）

#include <iostream>#include <cstdio>#include <cstring>#include <map>#include <set>#include <string>#include <cmath>#include <algorithm>#include <vector>#include <bitset>#include <stack>#include <queue>#include <functional>using namespace std;int R;int x[3],y[3];double ddabs(double xx){    if(xx < 0) return -xx;    return xx;}int main(){    int t;    scanf("%d",&t);    while(t--)    {        scanf("%d",&R);        for(int i=1;i<=2;i++) scanf("%d%d",&x[i],&y[i]);        int rr=(x[1]*x[1])+(y[1]*y[1]);        if(rr == 0)        {            printf("%.7f\n", R * 2.0);            continue;        }        int dd=(x[1]-x[2])*(x[1]-x[2])+(y[1]-y[2])*(y[1]-y[2]);        double o=acos((2.0*rr-dd)/rr/2.0);        double a,b,ans1;        double r=sqrt(1.0 * rr);        a=1.0*(R*R+rr)*cos(o/2.0)/(2.0*R*r);        b=a*a-(1.0+(2.0*rr-dd)/rr/2.0)/2.0;        if(ddabs(b) < 1e-8)            b = 0;        ans1=a+sqrt(b);        double t;        if(ans1 > 1 || ans1 < cos(o))        {            t = o/2.0;        }        else        {            t=acos(ans1)*2.0;            t=(t+o)/2.0;        }        double x=sqrt(R*R+rr-2.0*R*r*cos(t));        double y=sqrt(R*R+rr-2.0*R*r*cos(o-t));        printf("%.7f\n",x+y);    }    return 0;}