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-)  return ;
     return ;
 }

 Point bs(double x1,double y1,double x2,double y2,double x3,double y3,double R)  //二分查找交点
 {
     double l=,r=,mid,x,y;
     while(l<=r)
     {
         mid = (l+r)/;
         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 ;
     }
     double x1,y1;
     x1 = Px-Qx;
     y1 = Py-Qy;
     double x=,y=,k=,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\n",ans);
     return ;
 }