程式師世界 >> 編程語言 >> C語言 >> C++ >> C++入門知識 >> HDU 2857 Mirror and Light （計算幾何求對稱點和兩直線的交點）

HDU 2857 Mirror and Light （計算幾何求對稱點和兩直線的交點）

編輯：C++入門知識

題意：給你鏡子的位置（用兩點確定的一條直線表示），光源，光的反射點，求光在鏡子的折射點

計算幾何的模板，注意斜率！！！

模板1：

#include
#include
#include
#include
#include
#include
#include
#include 
#include 
#include
#include
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的對稱點
{
	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)//斜率為零
		{
			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)//求兩直線的交點
{
	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;
}

模板2（利用點到直線上的最近點避免斜率）：

#include
#include
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;//源點與對稱點的中點落在直線上
        now.y = 2*dot.y - l1.y;
        point ans = intersection(now,l2,m1,m2);
        printf("%.3lf %.3lf\n",ans.x,ans.y);
    }
    return 0;
}