[ACM] hdu 2857 Mirror and Light (对称点+两条直线的交点）

Problem Description

The light travels in a straight line and always goes in the minimal path between two points, are the basic laws of optics.

Now, our problem is that, if a branch of light goes into a large and infinite mirror, of course，it will reflect, and leave away the mirror in another direction. Giving you the position of mirror and the two points the light goes in before and after the reflection,
calculate the reflection point of the light on the mirror.

  

You can assume the mirror is a straight line and the given two points can’t be on the different sizes of the mirror.

Input

The first line is the number of test case t(t<=100).

  

The following every four lines are as follow:

  X1 Y1

  X2 Y2

  Xs Ys

  Xe Ye

  (X1,Y1),(X2,Y2) mean the different points on the mirror, and (Xs,Ys) means the point the light travel in before the reflection, and (Xe,Ye) is the point the light go after the reflection.

  The eight real number all are rounded to three digits after the decimal point, and the absolute values are no larger than 10000.0.

Output

  Each lines have two real number, rounded to three digits after the decimal point, representing the position of the reflection point.

Sample Input

1
0.000 0.000
4.000 0.000
1.000 1.000
3.000 1.000

Sample Output

2.000 0.000

Source

2009 Multi-University Training Contest 5 - Host by NUDT
 

模板：

const double eps=1e-6;
struct point
{
double x,y;
};
struct line//直线ax+by+c=0
{
double a,b,c;
};
line LineFromSegment(point p1,point p2)//两个点求直线
{
line temp;
temp.a=p2.y-p1.y;
temp.b=p1.x-p2.x;
temp.c=p2.x*p1.y-p1.x*p2.y;
return temp;
}
point LineInter(line l1,line l2)//两条直线求交点
{
point temp;
double a1=l1.a;
double b1=l1.b;
double c1=l1.c;
double a2=l2.a;
double b2=l2.b;
double c2=l2.c;
if(fabs(b1)<eps)
{
temp.x=-c1/a1;
temp.y=(-c2-a2*temp.x)/b2;
}
else
{
temp.x=(c1*b2-b1*c2)/(b1*a2-b2*a1);
temp.y=(-c1-a1*temp.x)/b1;
}

return temp;
}
point symmetrical(point p, line L)//求一个点关于一条直线的对称点
{
point p2;
double d;
d = L.a * L.a + L.b * L.b;
p2.x = (L.b * L.b * p.x - L.a * L.a * p.x -
2 * L.a * L.b * p.y - 2 * L.a * L.c) / d;
p2.y = (L.a * L.a * p.y - L.b * L.b * p.y -
2 * L.a * L.b * p.x - 2 * L.b * L.c) / d;
return p2;
}

本题代码：

#include <iostream>
#include <cmath>
#include <iomanip>
using namespace std;
const double eps=1e-6;
struct point
{
double x,y;
};
struct line//直线ax+by+c=0
{
double a,b,c;
};
line LineFromSegment(point p1,point p2)//两个点求直线
{
line temp;
temp.a=p2.y-p1.y;
temp.b=p1.x-p2.x;
temp.c=p2.x*p1.y-p1.x*p2.y;
return temp;
}
point LineInter(line l1,line l2)//两条直线求交点
{
point temp;
double a1=l1.a;
double b1=l1.b;
double c1=l1.c;
double a2=l2.a;
double b2=l2.b;
double c2=l2.c;
if(fabs(b1)<eps)
{
temp.x=-c1/a1;
temp.y=(-c2-a2*temp.x)/b2;
}
else
{
temp.x=(c1*b2-b1*c2)/(b1*a2-b2*a1);
temp.y=(-c1-a1*temp.x)/b1;
}

return temp;
}
point symmetrical(point p, line L)//求一个点关于一条直线的对称点
{
point p2;
double d;
d = L.a * L.a + L.b * L.b;
p2.x = (L.b * L.b * p.x - L.a * L.a * p.x -
2 * L.a * L.b * p.y - 2 * L.a * L.c) / d;
p2.y = (L.a * L.a * p.y - L.b * L.b * p.y -
2 * L.a * L.b * p.x - 2 * L.b * L.c) / d;
return p2;
}
int main()
{
int t;
cin>>t;
for(int i=1;i<=t;i++)
{
point p[4];
for(int j=0;j<4;j++)
cin>>p[j].x>>p[j].y;
point tmp;
line l1,l2;
l1=LineFromSegment(p[0],p[1]);//两点求直线
tmp=symmetrical(p[3],l1);//直线外一点与一条直线，求该点关于直线的对称点,这里是求的反射点的对称点
l2=LineFromSegment(tmp,p[2]);//对称点与入射点求直线
tmp=LineInter(l1,l2);//求反射点（两条直线相交的点）
cout<<setiosflags(ios::fixed)<<setprecision(3)<<tmp.x<<" "<<tmp.y<<endl;
}
return 0;
}

辅助图（根据所写代码画的）