HDU 2857 Mirror and Light(几何）

题目链接

简单求反射点。。。这种几何题，写成程序怎么这么容易错啊。4Y。。。中间各种小错误。。

#include <stdio.h>
#include <string.h>
#define eps 0.00000001
double xmi,ymi,xx,yy;
void line(double x1,double y1,double x2,double y2,double xs,double ys)//求（xs，ys）在直线上的对称点
{
    double k1,k2,b1,b2;
    if(y2 == y1)
    {
        xmi = xs;
        ymi = 2*y1 - ys;
    }
    else if(x2 == x1)
    {
        xmi = 2*x1 - xs;
        ymi = ys;
    }
    else
    {
        k1 = (y2-y1)/(x2-x1);
        k2 = (x1-x2)/(y2-y1);
        b1 = y2 - k1*x2;
        b2 = ys - k2*xs;
        xmi = 2*(b1-b2)/(k2-k1) - xs;//求对称点的x坐标
        ymi = k2*(xmi- xs) + ys;//带入垂直直线求y
    }
}
void  inst(double xs,double ys,double xe,double ye,double x1,double y1,double x2,double y2)//求两条直线的交点。
{
    double k1,k2,b1,b2;//两条直线不存在平行的情况
    if(xe == xs)
    {
        k2 = (y2-y1)/(x2-x1);
        b2 = y2 - k2*x2;
        xx = xe;
        yy = k2*xx + b2;
        return ;
    }
    if(x2 == x1)
    {
        k1 = (ye-ys)/(xe-xs);
        b1 = ys - k1*xs;
        xx = x2;
        yy = k1*xx + b1;
        return ;
    }
    k1 = (ye-ys)/(xe-xs);
    k2 = (y2-y1)/(x2-x1);
    b1 = ys - k1*xs;
    b2 = y2 - k2*x2;
    xx = (b1-b2)/(k2-k1);
    yy = k1*xx + b1;
}
int main()
{
    double x1,y1,x2,y2,xs,ys,xe,ye;
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lf%lf%lf%lf%lf%lf%lf%lf",&x1,&y1,&x2,&y2,&xs,&ys,&xe,&ye);
        line(x1,y1,x2,y2,xs,ys);
        inst(xmi,ymi,xe,ye,x1,y1,x2,y2);
        printf("%.3lf %.3lf\n",xx+eps,yy+eps);
    }
    return 0;
}
/*
查错的数据
2
1 1
0 0
0 1
-1 0
1 1 
0 0
0 1
1 2
*/