codeforces 935 C. Fifa and Fafa

最新推荐文章于 2019-08-05 21:04:32 发布

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本文探讨了一个有趣的问题：如何在一个圆形房间内放置一个圆形WiFi热点，使其覆盖面积最大，同时不超出房间边界且不覆盖特定禁止区域（如Fafa的所在位置）。文章提供了详细的数学解析方法及其实现代码。

Description

Fifa and Fafa are sharing a flat. Fifa loves video games and wants to download a new soccer game. Unfortunately, Fafa heavily uses the internet which consumes the quota. Fifa can access the internet through his Wi-Fi access point. This access point can be accessed within a range of r meters (this range can be chosen by Fifa) from its position. Fifa must put the access point inside the flat which has a circular shape of radius R. Fifa wants to minimize the area that is not covered by the access point inside the flat without letting Fafa or anyone outside the flat to get access to the internet.

The world is represented as an infinite 2D plane. The flat is centered at (x1, y1) and has radius R and Fafa’s laptop is located at (x2, y2), not necessarily inside the flat. Find the position and the radius chosen by Fifa for his access point which minimizes the uncovered area.

Input

The single line of the input contains 5 space-separated integers R, x1, y1, x2, y2 (1 ≤ R ≤ 10 $^5$ , |x1|, |y1|, |x2|, |y2| ≤ 10 $^5$ ).

Output

Print three space-separated numbers x $_{ap}$ , y $_{ap}$ , r where (x $_{ap}$ , y $_{ap}$ ) is the position which Fifa chose for the access point and r is the radius of its range.

Your answer will be considered correct if the radius does not differ from optimal more than 10 $^{- 6}$ absolutely or relatively, and also the radius you printed can be changed by no more than 10 $^{ - 6}$ (absolutely or relatively) in such a way that all points outside the flat and Fafa’s laptop position are outside circle of the access point range.

Examples

Input
5 3 3 1 1
Output
3.7677669529663684 3.7677669529663684 3.914213562373095

Input
10 5 5 5 15
Output
5.0 5.0 10.0

题意：在一个圆形房间中，设置一个最大的圆形WI-FI范围，该范围不能超出房间，也不能覆盖fafa所在的点。
思路：先以房间中心建立相对坐标系，再选取fafa所在的直径长的那部分作为WI-FI的直径。

sketch

如图，(x3, y3)利用相似三角形是容易计算的，x3= $\frac{-x_1R}{\sqrt{x_1^2+y_1^2}}$ ，y3= $\frac{-y_1R}{\sqrt{x_1^2+y_1^2}}$ ，所以x2= $\frac{x_1+x_3}{2}$ ，y2= $\frac{y_1+y_3}{2}$ ，半径为 $\frac{R+R_a}{2}$

#include<stdio.h>
#include<math.h>
int main(){
   double R,aR,iR,xR,yR,xa,ya,xi,yi,xt,yt;
    scanf("%lf %lf %lf %lf %lf",&R,&xR,&yR,&xa,&ya);
    xa=xa-xR;
    ya=ya-yR;
    //相对坐标
    aR=sqrt(xa*xa+ya*ya);
    //Fafa在外面的情况
    if(aR-R>0.0000001){
        printf("%.10f %.10f %.10f\n",xR,yR,R);
        return 0;
    }
    //Fafa在中心的情况
    if(aR<0.0000001){
       printf("%.10f %.10f %.10f\n",xR+0.5*R,yR,R*0.5);
        return 0;
    }
    xt=-1*xa*R/aR;
    yt=-1*ya*R/aR;
    xi=(xt+xa)/2;
    yi=(yt+ya)/2;
    iR=(R+aR)/2;
    printf("%.10f %.10f %.10f\n",xi+xR,yi+yR,iR);
    return 0;
}