(模板)计算几何点线面形基础知识总结

  计算几何点线面形基础知识总结模板:

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cstdlib>
using namespace std;

const double eps=1e-8;
const double inf=1e20;int sgn(double x){
    if(abs(x)<eps) return 0;
    if(x<0) return -1;
    return 1;
}

struct Point{
    double x,y;
    Point(double xx=0,double yy=0):x(xx),y(yy){}
    bool operator == (const Point& b)const{
        sgn(x-b.x)==0&&sgn(y-b.y)==0;
    }
    Point operator + (const Point& b)const{
        return Point(x+b.x,y+b.y);
    }
    Point operator - (const Point& b)const{
        return Point(x-b.x,y-b.y);
    }
    Point operator * (const double& k)const{  //返回相乘后的点
        return Point(x*k,y*k);
    }
    Point operator / (const double& k)const{  //返回相除后的点
        return Point(x/k,y/k);
    }
    double operator * (const Point& b)const{  //点乘
        return x*b.x+y*b.y;
    }
    double operator ^ (const Point& b)const{  //叉乘
        return x*b.y-b.x*y;
    }
    double len(){    //返回长度
        return sqrt(x*x+y*y);
    }
    double len2(){   //返回长度的平方
        return x*x+y*y;
    }
    Point norm(double r){  //标准化,返回将长度化为r的点
        double l=len();
        if(!sgn(l)) return *this;
        r/=l;
        return Point(x*r,y*r);
    }
    //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;
    Line(){}
    Line(Point ss,Point ee){
        s=ss,e=ee;
    }
    //返回点p在直线上的投影
    Point PointProg(Point p){
        return s+(((e-s)*((e-s)*(p-s)))/((e-s).len2()));
    }
    //返回点p关于直线的对称点
    Point PointSymmetry(Point p){
        Point q=PointProg(p);
        return Point(2*q.x-p.x,2*q.y-p.y);
    }
    //点是否在直线上
    bool PointOnLine(Point p){
        return sgn((p-s)^(e-s))==0;
    }
    //点是否在线段上
    bool PointOnSeg(Point p){
        return sgn((p-s)^(e-s))==0&&sgn((p-s)*(p-e))<=0;
    }
    //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2表示相交
    //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);
    }
};

struct Circle{
    Point p;
    double r;
    Circle(){}
    Circle(Point pp,double rr){
        p=pp,r=rr;
    }
    //求直线与圆的交点,返回交点个数
    int LineCrossCircle1(Line l,Point &p1,Point &p2){
        double dx=l.e.x-l.s.x,dy=l.e.y-l.s.y;
        double a=dx*dx+dy*dy;
        double b=2*dx*(l.s.x-p.x)+2*dy*(l.s.y-p.y);
        double c=(l.s.x-p.x)*(l.s.x-p.x)+(l.s.y-p.y)*(l.s.y-p.y)-r*r;
        double d=b*b-4*a*c;
        if(sgn(d)<0) return 0;
        int res=0;
        double ans1=(-b-sqrt(d))/(2.0*a);
        double ans2=(-b+sqrt(d))/(2.0*a);
        
        p1=Point(l.s.x+ans1*dx,l.s.y+ans1*dy);
        ++res;
        p2=Point(l.s.x+ans2*dx,l.s.y+ans2*dy);
        ++res;
    
        if(res==2&&sgn(ans1-ans2)==0) --res;
        return res;
    }
    //求射线与圆的交点,返回交点个数
    int LineCrossCircle2(Line l,Point &p1,Point &p2){
        double dx=l.e.x-l.s.x,dy=l.e.y-l.s.y;
        double a=dx*dx+dy*dy;
        double b=2*dx*(l.s.x-p.x)+2*dy*(l.s.y-p.y);
        double c=(l.s.x-p.x)*(l.s.x-p.x)+(l.s.y-p.y)*(l.s.y-p.y)-r*r;
        double d=b*b-4*a*c;
        if(sgn(d)<0) return 0;
        int res=0;
        double ans1=(-b-sqrt(d))/(2.0*a);
        double ans2=(-b+sqrt(d))/(2.0*a);
        if(sgn(ans1)>=0){
            p1=Point(l.s.x+ans1*dx,l.s.y+ans1*dy);
            ++res;
        }
        if(sgn(ans2)>=0){
            p2=Point(l.s.x+ans2*dx,l.s.y+ans2*dy);
            ++res;
        }
        if(res==2&&sgn(ans1-ans2)==0) --res;
        return res;
    }
};

//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((l1.s-l2.s)^(l2.e-l2.s))*sgn((l1.e-l2.s)^(l2.e-l2.s))<=0&&
        sgn((l2.s-l1.s)^(l1.e-l1.s))*sgn((l2.e-l1.s)^(l1.e-l1.s))<=0;
}

double dis(Point a,Point b){
    return sqrt((b-a)*(b-a));
}
//判断点在线段上
bool OnSeg(Point P,Line L){
    return
        sgn((L.s-P)^(L.e-P))==0&&
        sgn((P.x-L.s.x)*(P.x-L.e.x))<=0&&
        sgn((P.y-L.s.y)*(P.y-L.e.y))<=0;
}
//判断点在凸多边形内,复杂度O(n)
//点形成一个凸包,而且按逆时针排序(如果是顺时针把里面的<0改为>0)
//点的编号:0~n-1
//返回值:
//-1:点在凸多边形外
//0:点在凸多边形边界上
//1:点在凸多边形内
int inConvexPoly(Point a,Point p[],int n){
    for(int i=0;i<n;++i)
        if(sgn((p[i]-a)^(p[(i+1)%n]-a))<0) return -1;
        else if(OnSeg(a,Line(p[i],p[(i+1)%n]))) return 0;
    return 1;
}
//判断点在任意多边形内,复杂度O(n)
//射线法,poly[]的顶点数要大于等于3,点的编号0~n-1,按逆时针或顺时针排序
//返回值
//-1:点在凸多边形外
//0:点在凸多边形边界上
//1:点在凸多边形内
int inPoly(Point a,Point p[],int n){
    int cnt=0;
    Line ray,side;
    ray.s=a;
    ray.e.y=a.y;
    ray.e.x=-inf;
    for(int i=0;i<n;++i){
        side.s=p[i];
        side.e=p[(i+1)%n];
        if(OnSeg(a,side)) return 0;
        if(sgn(side.s.y-side.e.y)==0) continue;
        if(OnSeg(side.s,ray)){
            if(sgn(side.s.y-side.e.y)>0) ++cnt;
        }
        else if(OnSeg(side.e,ray)){
            if(sgn(side.e.y-side.s.y)>0) ++cnt;
        }
        else if(inter(ray,side)) ++cnt;
    }
    if(cnt%2==1) return 1;
    else return -1;
}
//判断凸多边形
//允许共线边
//点可以是顺时针给出也可以是逆时针给出
//点的编号是0~n-1
bool isconvex(Point poly[],int n){
    bool s[3];
    memset(s,false,sizeof(s));
    for(int i=0;i<n;++i){
        s[sgn((poly[(i+1)%n]-poly[i])^(poly[(i+2)%n]-poly[i]))+1]=true;
        if(s[0]&&s[2]) return false;
    }
    return true;
}
//点到线段的距离
//返回点到线段最近的点
Point NearestPointToLineSeg(Point P,Line L){
    Point result;
    double t=((P-L.s)*(L.e-L.s))/((L.e-L.s)*(L.e-L.s));
    if(t>=0&&t<=1){
        result.x=L.s.x+(L.e.x-L.s.x)*t;
        result.y=L.s.y+(L.e.y-L.s.y)*t;
    }
    else{
        if(dis(P,L.s)<dis(P,L.e))
            return L.s;
        else 
            return L.e;
    }
    return result;
}

以pt点为中心进行极角排序:

struct Point{
    int id,x,y;
    Point(){}
    Point(int x,int y):x(x),y(y){}
}p1[maxn],p2[maxn],pt;

Point operator - (const Point& a,const Point& b){
    return Point(a.x-b.x,a.y-b.y);
}

bool operator ^ (const Point& a,const Point& b){
    return (1LL*a.x*b.y-1LL*b.x*a.y)>0;
}
//返回向限
inline int Quadrant(const Point& a){
    if(a.x>0&&a.y>=0) return 1;
    else if(a.x<=0&&a.y>0) return 2;
    else if(a.y<=0&&a.x<0) return 3;
    else if(a.x>=0&&a.y<0) return 4;
}
//以pt为中心进行极角排序
bool operator < (const Point& c,const Point& d){
    Point a=c-pt,b=d-pt;
    int qa=Quadrant(a),qb=Quadrant(b);
    if(qa!=qb) return qa<qb;
    return a^b;
}

(其中叉积求直线的交点的证明见https://www.cnblogs.com/jklover/p/10484313.html,叉积判断线段相交用到了快速排斥实验和跨立实验。)

判断线段是否在多边形内:当多边形为凸时,只需判断线段的两个端点在多边形内即可。

            当多边形为凹时,其伪代码如下,复杂度为O(n):

              

原文地址:https://www.cnblogs.com/FrankChen831X/p/11498774.html