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  • Codeforces 32E Hide-and-Seek 乞讨2关于镜面反射点 计算几何

    主题链接:点击打开链接

    必须指出的是,反射镜和2个人共线是不是障碍,但根据该壁其他情况

    #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)
    ",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
    //仅仅有第一个值为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));}
    //*推断线段相交
    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; }


    版权声明:本文博主原创文章,博客,未经同意不得转载。

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  • 原文地址:https://www.cnblogs.com/blfshiye/p/4854527.html
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