HDU 2857 Mirror and Light （计算几何求 对称点和两直线的交点）

#include<cstdio>
#include<stdlib.h>
#include<string.h>
#include<string>
#include<map>
#include<cmath>
#include<iostream>
#include <queue>
#include <stack>
#include<algorithm>
#include<set>
using namespace std;
#define INF 1e8
#define eps 1e-4
#define ll __int64
#define maxn 500010
#define mol 1000000007

struct point
{
double x,y;
};
point symmetric_point(point p1, point l1, point l2)// 求p1的对称点
{
point ret;
if (l1.x > l2.x - eps && l1.x < l2.x + eps)//斜率不存在
{
ret.x = (2 * l1.x - p1.x);
ret.y = p1.y;
}
else
{
double k = (l1.y - l2.y ) / (l1.x - l2.x);
if(k + eps > 0 && k - eps < 0)//斜率为零
{
ret.x = p1.x;
ret.y = l1.y - (p1.y - l1.y);
}
else
{
ret.x = (2*k*k*l1.x + 2*k*p1.y - 2*k*l1.y - k*k*p1.x + p1.x) / (1 + k*k);
ret.y = p1.y - (ret.x - p1.x ) / k;
}
}
return ret;
}
point intersection(point u1,point u2,point v1,point v2)//求两直线的交点
{
point ret=u1;
double t=((u1.x-v1.x)*(v1.y-v2.y)-(u1.y-v1.y)*(v1.x-v2.x))
/((u1.x-u2.x)*(v1.y-v2.y)-(u1.y-u2.y)*(v1.x-v2.x));
ret.x+=(u2.x-u1.x)*t;
ret.y+=(u2.y-u1.y)*t;
return ret;
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
point m1,m2,l1,l2;
scanf("%lf%lf%lf%lf%lf%lf%lf%lf",&m1.x,&m1.y,&m2.x,&m2.y,&l1.x,&l1.y,&l2.x,&l2.y);
point tmp=symmetric_point(l1,m1,m2);
point ans=intersection(tmp,l2,m1,m2);
printf("%.3lf %.3lf\n",ans.x,ans.y);
}
return 0;
}

#include<cstdio>
#include<cmath>
using namespace std;
const double eps=1e-8;

struct point{
double x,y;
};

point intersection(point u1,point u2,point v1,point v2)
{
point ret=u1;
double t=((u1.x-v1.x)*(v1.y-v2.y)-(u1.y-v1.y)*(v1.x-v2.x))
/((u1.x-u2.x)*(v1.y-v2.y)-(u1.y-u2.y)*(v1.x-v2.x));
ret.x+=(u2.x-u1.x)*t;
ret.y+=(u2.y-u1.y)*t;
return ret;
}

point ptoline(point p,point l1,point l2)
{
point t=p;
t.x+=l1.y-l2.y,t.y+=l2.x-l1.x;
return intersection(p,t,l1,l2);
}

point m1,m2,l1,l2;

int main()
{
int n;
scanf("%d",&n);
while(n--){
scanf("%lf%lf%lf%lf%lf%lf%lf%lf",&m1.x,&m1.y,&m2.x,&m2.y,&l1.x,&l1.y,&l2.x,&l2.y);
point dot = ptoline(l1,m1,m2);
point now;
now.x = 2*dot.x - l1.x;//源点与对称点的中点落在直线上
now.y = 2*dot.y - l1.y;
point ans = intersection(now,l2,m1,m2);
printf("%.3lf %.3lf\n",ans.x,ans.y);
}
return 0;
}