Codeforces 32E Hide-and-Seek 求2点关于镜面反射 计算几何

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题目链接：点击打开链接

需要注意的是镜子在与2个人共线时是不作为障碍物，但其他情况与墙一致

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<string.h>
#include<math.h>
using namespace std;
#define point Point
const double eps = 1e-8;
const double PI = acos(-1.0);
double ABS(double x){return x>0?x:-x;}
int sgn(double x){
	if(fabs(x) < eps)return 0;
	if(x < 0)return -1;
	else return 1;
}
struct Point
{
	double x,y;
	void put(){printf("(%.0lf,%.0lf)\n",x,y);}
	Point(){}
	Point(double _x,double _y){
	x = _x;y = _y;
	}
	Point operator -(const Point &b)const{
		return Point(x - b.x,y - b.y);
	}
	//叉积
	double operator ^(const Point &b)const{
		return x*b.y - y*b.x;
	}
	//点积
	double operator *(const Point &b)const{
		return x*b.x + y*b.y;
	}
	//绕原点旋转角度B（弧度值），后x,y的变化
	void transXY(double B){
		double tx = x,ty = y;
		x = tx*cos(B) - ty*sin(B);
		y = tx*sin(B) + ty*cos(B);
	}
};
struct Line
{
Point s,e;
void put(){s.put();e.put();}
Line(){}
Line(Point _s,Point _e)
{
s = _s;e = _e;
}
//两直线相交求交点
//第一个值为0表示直线重合，为1表示平行，为0表示相交,为2是相交
//只有第一个值为2时，交点才有意义
pair<int,Point> operator &(const Line &b)const{
	Point res = s;
	if(sgn((s-e)^(b.s-b.e)) == 0)
	{
	if(sgn((s-b.e)^(b.s-b.e)) == 0)
	return make_pair(0,res);//重合
	else return make_pair(1,res);//平行
	}
	double t = ((s-b.s)^(b.s-b.e))/((s-e)^(b.s-b.e));
	res.x += (e.x-s.x)*t;
	res.y += (e.y-s.y)*t;
	return make_pair(2,res);
}
};
double dist(Point a,Point b){return sqrt((a-b)*(a-b));}
//*判断线段相交
bool inter(Line l1,Line l2)
{
	return
	max(l1.s.x,l1.e.x) >= min(l2.s.x,l2.e.x) &&
	max(l2.s.x,l2.e.x) >= min(l1.s.x,l1.e.x) &&
	max(l1.s.y,l1.e.y) >= min(l2.s.y,l2.e.y) &&
	max(l2.s.y,l2.e.y) >= min(l1.s.y,l1.e.y) &&
	sgn((l2.s-l1.e)^(l1.s-l1.e))*sgn((l2.e-l1.e)^(l1.s-l1.e)) <= 0 &&
	sgn((l1.s-l2.e)^(l2.s-l2.e))*sgn((l1.e-l2.e)^(l2.s-l2.e)) <= 0;
}
point symmetric_point(point p1, point l1, point l2)
{
	point ret;
	if(ABS(l1.x-l2.x)<eps){
		ret.y = p1.y;
		ret.x = 2*l1.x - p1.x;
		return ret;
	}
	if(ABS(l1.y-l2.y)<eps) {
		ret.x = p1.x;
		ret.y = 2*l1.y - p1.y;
		return ret;
	}
	if (l1.x > l2.x - eps && l1.x < l2.x + eps)
	{
	ret.x = (2 * l1.x - p1.x);
	ret.y = p1.y;
	}
	else
	{
	double k = (l1.y - l2.y ) / (l1.x - l2.x);
	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;
}
bool gongxian(Point a, Point b, Point c){
	return ABS((a.y-b.y)*(a.x-c.x) - (a.y-c.y)*(a.x-b.x))<eps;
}
Point a,b;
Line peo, wal, mir;
bool work(){
//	peo.put(); wal.put();
	if((gongxian(a,mir.s,mir.e)&&gongxian(b,mir.s,mir.e))) {
	//	a.put(); b.put(); mir.put();		puts("共线了");
		return !inter(wal,peo);
	}
	else if(inter(mir,peo))return false;
	if(!inter(wal, peo)) return true;
//	puts("GEGE");
	if(inter(wal,mir))return false;
	Point c = symmetric_point(b,mir.s,mir.e);
	Point d = symmetric_point(a,mir.s,mir.e);
//	c.put(); d.put();
	Line ac, bd;
	ac.s = a, ac.e = c;
	bd.s = b, bd.e = d;
	if(!inter(ac,mir))return false;
	if(!inter(bd,mir))return false;
	if(inter(ac,wal))return false;
	if(inter(bd,wal))return false;
	return true;
}
int main(){
    while(~scanf("%lf %lf",&a.x,&a.y)){
		scanf("%lf %lf",&b.x,&b.y);
		scanf("%lf %lf %lf %lf", &wal.s.x, &wal.s.y, &wal.e.x, &wal.e.y);
		scanf("%lf %lf %lf %lf", &mir.s.x, &mir.s.y, &mir.e.x, &mir.e.y);
		peo.s = a, peo.e = b;
		work()?puts("YES"):puts("NO");
    }
    return 0;
}

Codeforces 32E Hide-and-Seek 求2点关于镜面反射 计算几何,布布扣,bubuko.com

Codeforces 32E Hide-and-Seek 求2点关于镜面反射 计算几何

标签：http   os   for   io   line   amp   

原文地址：http://blog.csdn.net/qq574857122/article/details/37600999