没事做，贴个代码。

Problem E

Time Limit: 2000/1000 MS (Java/Others)     Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11    Accepted Submission(s): 2

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.

 

SampleInput

1
0.000 0.000
4.000 0.000
1.000 1.000
3.000 1.000

 

SampleOutput

2.000 0.000 

 

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

#define abs(double) ((double)>0?(double):-(double))
#define eps 0.0000001

using namespace std;

struct point{
 double x,y;
 friend istream&operator>>(istream&in,point &p){
  in>>p.x>>p.y;
  return in;
 }
 friend ostream&operator<<(ostream&out,point p){
  out<<p.x<<' '<<p.y<<endl;
  return out;
 }
};
struct line{
 bool inf;
 double k,b;
 friend istream&operator>>(istream&in,line&l){
  l.inf=false;
  in>>l.k>>l.b;
  return in;
 }
 friend ostream&operator<<(ostream&out,line l){
  if(l.inf)out<<' '<<l.b<<endl;
  else out<<l.k<<' '<<l.b<<endl;
  return out;
 }
 friend point operator*(line l1,line l2){
  point t;
  if(l1.inf){
   t.x=l1.b;
   t.y=l2.k*t.x+l2.b;
  }else if(l2.inf){
   t.x=l2.b;
   t.y=l1.k*t.x+l1.b;
  }else{
   t.x=(l2.b-l1.b)/(l1.k-l2.k);
   t.y=l1.k*t.x+l1.b;
  }
  return t;
 }
};
line point2line(point a,point b){
 line t;
 t.inf=abs(a.x-b.x)<eps;
 if(t.inf){
  t.b=a.x;
 }else{
  t.k=(a.y-b.y)/(a.x-b.x);
  t.b=a.y-t.k*a.x;
 }
 return t;
}
point pointcalc(point p,line l){
 point t;
 if(l.inf){
  t.x=l.b*2-p.x;
  t.y=p.y;
 }else{
  t.x=((1-l.k*l.k)*p.x+2*l.k*p.y-2*l.k*l.b)/(1+l.k*l.k);
  t.y=((l.k*l.k-1)*p.y+2*l.k*p.x+2*l.b)/(1+l.k*l.k);
 }
 return t;
}
int main(){
 int cas;
 line t1,t2;
 point a,b,c,d,x;
 cin>>cas;
 while(cas--){
  cin>>a>>b>>c>>d;
  x=pointcalc(c,t1=point2line(a,b));
  //cout<<t1<<x<<d;
  t2=point2line(x,d);
  //cout<<t2;
  c=t1*t2;
  printf("%.3f %.3f\n",c.x,c.y);
 }
 return 0;
}