hdu 2857 Mirror and Light 求两直线交点

最新推荐文章于 2018-09-22 21:59:24 发布

原创最新推荐文章于 2018-09-22 21:59:24 发布 · 778 阅读

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本文提供了一种解决两直线交点问题的算法实现，并通过具体代码示例展示了如何根据不同情况（如垂直线和平行线）高效计算交点坐标。

hdu 2857 Mirror and Light 两直线交点

#include <iostream>

#include <cstring>
#include <cstdio>
#include <cmath>
#define eps 1e-8
using namespace std;
double x[5],y[5],k1;
void qiushuzhi()
{
double k2;
k2=(y[4]-y[3])/(x[4]-x[3]);
printf("%.3lf %.3lf\n",x[1],k2*(x[1]-x[3])+y[3]);
}
void qiushuipin()
{
double k2;
k2=(y[4]-y[3])/(x[4]-x[3]);
printf("%.3lf %.3lf\n",(y[1]-y[3])/k2+x[3],y[1]);
}
void qiucrosspoint()
{ double k2,a,b;
if(fabs(x[3]-x[4])<eps)
{
printf("%.3lf %.3lf\n",x[3],k1*(x[3]-x[1])+y[1]);
return ;
}
k2=(y[4]-y[3])/(x[4]-x[3]);
if(fabs(k1-k2)<eps)return;
a=(y[3]-y[1]+k1*x[1]-k2*x[3])/(k1-k2);
b=k1*(a-x[1])+y[1];
printf("%.3lf %.3lf\n",a,b);
}
int main()
{
int t;
double xzh,yzh;
scanf("%d",&t);
while(t--)
{
scanf("%lf%lf%lf%lf",&x[1],&y[1],&x[2],&y[2]);
scanf("%lf%lf%lf%lf",&x[3],&y[3],&x[4],&y[4]);
if(fabs(x[1]-x[2])<eps)
{
x[3]=-x[3];
qiushuzhi();
continue;
}
if(fabs(y[1]-y[2])<eps)
{
y[3]=-y[3];
qiushuipin();
continue;
}
k1=(y[2]-y[1])/(x[2]-x[1]);
xzh=(k1*y[3]-k1*y[1]+k1*k1*x[1]+x[3])/(k1*k1+1);
yzh=k1*(xzh-x[1])+y[1];
x[3]=2*xzh-x[3];
y[3]=2*yzh-y[3];
qiucrosspoint();
}
return 0;

}

较简洁的代码:

#include<stdio.h>

int main()

{

   int t;

   double x1,y1,x2,y2,xs,ys,xe,ye;

   double k1,k2,x3,y3,a,b;

   scanf("%d",&t);

   while(t--)

   {

       scanf("%lf%lf%lf%lf%lf%lf%lf%lf",&x1,&y1,&x2,&y2,&xs,&ys,&xe,&ye);

       if(x1==x2)

       {

           y3=ys;

           x3=(-1)*xs;

           k2=(ye-y3)/(xe-x3);

           a=0;

           b=ye-k2*xe;

       }

       else

       {

           k1=(y1-y2)/(x1-x2);

           y3=(k1*k1*ys-ys+2*xs*k1+2*y2-2*k1*x2)/(1+k1*k1);

           x3=xs-k1*y3+k1*ys;

           if(xe==x3)

           {

               a=0;

               b=y2-k1*x2;

           }

           else

           {

               k2=(ye-y3)/(xe-x3);

               a=(ye-y2+k1*x2-k2*xe)/(k1-k2);

               b=k1*a-k1*x2+y2;

           }

       }

       printf("%.3f %.3f\n",a,b);

   }

   return 0;

}