HDU 6097---Mindis(二分)

2018-06-17 22:04:53来源:未知 阅读 ()

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题目链接

 

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
T500000
100x,y100
1r100
 

 

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 106.
Formally, let your answer be a, and the jury's answer be b. Your answer is considered correct if |ab|max(1,b)106.
 

 

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
 
题意:有一个圆,圆心在原点,输入半径 r ,圆内有两个点P , Q 到圆心距离相同,现在求在圆上找一点M,使得PM+QM最小?
 
思路:
            
      对于圆内的两个点 P 和 Q ,如果以其作为椭圆两个焦点,由图一可以看到:当椭圆顶点与圆相切时 , 圆与椭圆会有三个焦点,而椭圆上的点到P、Q的距离和是定值。那么可以缩小椭圆,接下来会有四个交点,继续缩小椭圆,得到图二,椭圆与圆只有两个交点,因为椭圆外的点到两个焦点的距离和大于2a,而椭圆上的点到两个焦点的距离和为2a,。在图2中,除了两个交点外,其余圆上点均在椭圆外,所以这时的交点到P、Q两点的距离和最小,等于2a。
      如何找到这个相切的椭圆呢?二分解决。
      具体:由P、Q两个焦点,我们可以确定c=PQ/2 ,  现在如果再给出一个b值那么就能确定这个椭圆了,而b值越大椭圆越大,b值越小椭圆越小,所以我们要找到如图2所示,相切只有两个焦点时的b值,那么我们需要找的就是圆和椭圆相交时最小的b。我们可以求出图3中所示的 h 和 R -h ,那么椭圆的方程为:x^2/a^2 + (y-h)^2/b^2 = 1 ( 这个方程是PQ平行于X轴时的,但PQ不平行于X时也不影响,我们只是用它判断是否与圆相交 ) 圆:x^2 + y^2 =R^R ,    并且b>0(显然),b<R-h ,所以在(0,R-h) 二分查找b。
 
代码如下:
#include <iostream>
#include <stdio.h>
#include <math.h>
#include <algorithm>
using namespace std;
const double eps = 1e-10;
double dis(double x1,double y1,double x2,double y2 )
{
    return sqrt((x2-x1)*(x2-x1)+(y2-y1)*(y2-y1));
}
namespace IO {
    const int MX = 4e7; //1e7占用内存11000kb
    char buf[MX]; int c, sz;
    void begin() {
        c = 0;
        sz = fread(buf, 1, MX, stdin);
    }
    inline bool read(int &t) {
        while(c < sz && buf[c] != '-' && (buf[c] < '0' || buf[c] > '9')) c++;
        if(c >= sz) return false;
        bool flag = 0; if(buf[c] == '-') flag = 1, c++;
        for(t = 0; c < sz && '0' <= buf[c] && buf[c] <= '9'; c++) t = t * 10 + buf[c] - '0';
        if(flag) t = -t;
        return true;
    }
}

int main()
{
    IO::begin();
    int T;
    double R,x1,x2,y1,y2;
    double x3,y3;
    double A,B,C,dt,ans1,ans2;
    //cin>>T;
    IO::read(T);
    while(T--)
    {
        //scanf("%lf%lf%lf%lf%lf",&R,&x1,&y1,&x2,&y2);
        int xr, xx1, xx2, yy1, yy2;
        IO::read(xr);
        IO::read(xx1);
        IO::read(yy1);
        IO::read(xx2);
        IO::read(yy2);
        R = xr;
        x1 = xx1;y1 = yy1;x2 = xx2;y2 = yy2;
        double c=dis(x1,y1,x2,y2)*0.5;
        c=c*c;
        x3=(x1+x2)*0.5;
        y3=(y1+y2)*0.5;
        double h=dis(x3,y3,0.0,0.0);
        //cout<<h<<endl;
        double Rb=R-h;
        double Lb=0.0,b,a,mid;

        for(int i=0; i<30; i++)
        {
            mid=(Lb+Rb)*0.5;
            b = mid;
            b=b*b;
            a=c+b;
            A=a-b;
            B=2.0*a*h;
            C=b*R*R+a*h*h-a*b;
            dt=B*B-4.0*A*C;
            int flag=1;
            if(dt>=-eps)
            {
                ans1=(-B+sqrt(B*B-4.0*A*C))*0.5/A;
                ans2=(-B-sqrt(B*B-4.0*A*C))*0.5/A;
                ans1=ans1*ans1;
                ans2=ans2*ans2;
                if(R*R-ans1>=-eps||R*R-ans2>=-eps)
                    flag=0;
            }
            if(flag==0)
                Rb=mid;
            else
                Lb=mid;
        }
        printf("%.10f\n",sqrt(a)*2.0);

    }
    return 0;
}

 

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