bnuoj 1053 EASY Problem (计算几何)

http://www.bnuoj.com/bnuoj/problem_show.php?pid=1053

【题意】：基本上就是求直线与圆的交点坐标

【题解】：这种题我都比较喜欢用二分，三分做，果然可以完爆，哈哈，特有成就感的说。。。

【code】：

#include <iostream>
#include <stdio.h>
#include <math.h>
#include <algorithm>

using namespace std;
#define eps 1e-12

struct Point
{
    double x,y;
    Point(){}
    Point(double dx,double dy)
    {
        x = dx;
        y = dy;
    }
};

double distance(double x1,double y1,double x2,double y2) //求距离
{
    return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
}

double distance2(double x1,double y1,double x2,double y2) //求距离的平方
{
    return ((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
}

double area2(double x1, double y1, double x2, double y2, double x3,double y3)  //两倍三角形面积
{
    double area;
    area = fabs(x1*y2+x2*y3+x3*y1-x3*y2-x1*y3-x2*y1);
    return area;
}

int isEqual(double a,double b)  //判断两浮点数是否相等
{
    if(fabs(a-b)>1e-8)  return 0;
    return 1;
}

Point bs(double x1,double y1,double x2,double y2,double x3,double y3,double R)  //二分查找交点
{
    double l=0,r=1,mid,x,y;
    while(l<=r)
    {
        mid = (l+r)/2;
        x = x1 + mid*(x2-x1);
        y = y1 + mid*(y2-y1);
        double temp =distance(x,y,x3,y3);
        if(temp<R) //与半径的距离为二分点
        {
            l=mid+eps;
        }
        else
        {
            r=mid-eps;
        }
    }
    return Point(x,y);
}

int main()
{
    double Cx,Cy,R;
    double Px,Py;
    double Qx,Qy;
    scanf("%lf%lf%lf",&Cx,&Cy,&R);
    scanf("%lf%lf",&Px,&Py);
    scanf("%lf%lf",&Qx,&Qy);
    double area = area2(Px,Py,Qx,Qy,Cx,Cy);
    double disPQ = distance(Px,Py,Qx,Qy);
    double dis = area/disPQ;  //用面积除以底求得三角形的高，即点到直线的距离
    if(dis>R||isEqual(dis,R))  
    {
        puts("-1");
        return 0;
    }
    double x1,y1;
    x1 = Px-Qx;
    y1 = Py-Qy;
    double x=0,y=0,k=0,lx,rx,ry,ly;
    if(isEqual(Px,Qx))  //如果px==qx，不存在斜率
    {
        x = Px;
        y = (Cx-x)*x1/y1+Cy;
        lx = x;
        ly = Cy-R;
        rx = x;
        ry = Cy+R;
    }
    else if(isEqual(Py,Qy))  //存在斜率为1
    {
        y = Py;
        x = (Cy-y)*y1/x1+Cx;
        ly = Py;
        lx = Px-R;
        ry = Py;
        rx = Px+R;
    }
    else //斜率在0-1之间
    {
        y = (Cy*y1/x1+Cx-Px+Py*(Qx-Px)/(Qy-Py))/(y1/x1+(Qx-Px)/(Qy-Py));
        x = (Cy-y)*y1/x1+Cx;
        lx = Cx - R;
        ly = (Qy-Py)/(Qx-Px)*(lx-Px)+Py;
        rx = Cx + R;
        ry = (Qy-Py)/(Qx-Px)*(rx-Px)+Py;
    }
    Point p1 = bs(x,y,lx,ly,Cx,Cy,R);
    Point p2 = bs(x,y,rx,ry,Cx,Cy,R);
    double ans = distance2(p1.x,p1.y,Px,Py)+distance2(p2.x,p2.y,Px,Py);
    printf("%.3lf
",ans);
    return 0;
}