• Potato的暑期训练day#2 ——计算几何模板


    计算几何

    二维几何:

    点与向量

    const double eps=1e-10;
    const double PI=acos(-1.0);
    
    struct Point{
        double x,y;
        Point(double x=0,double y=0):x(x),y(y){}
    };
    typedef Point Vector;
    
    Vector operator -(Point a,Point b){
        return Vector(a.x-b.x,a.y-b.y);
    }
    Vector operator +(Point a,Point b){
        return Vector(a.x+b.x,a.y+b.y);
    }
    Vector operator *(Vector a,double p){
        return Vector(a.x*p,a.y*p);
    }
    Vector operator /(Vector a,double p){
        return Vector(a.x/p,a.y/p);
    }
    bool operator <(const Point& a,const Point& b){
        return a.x<b.x||(a.x==b.x&&a.y<b.y);//在有精度需求,比如使用lower_bound的时候,加上dcmp()
    }
    int dcmp(double x){
        if(fabs(x)<eps)return 0;
        if(x<0)return -1;
        return 1;
    }
    bool operator ==(const Point& a,const Point& b){
        return dcmp(a.x-b.x)==0&&dcmp(a.y-b.y)==0;
    }
    double Dot(Vector a,Vector b){
        return a.x*b.x+a.y*b.y;
    }//点积
    double Cross(Vector a,Vector b){
        return a.x*b.y-a.y*b.x;
    }//叉积   
    double Length(Vector a){
        return sqrt(Dot(a,a));
    }//长度
    //返回逆时针旋转90度的单位法向量;
    Vector Normal(Vector a){
        double l=Length(a);
        return Vector(-a.y/l,a.x/l);
    }
    //返回向量夹角,无方向
    double Angle(Vector a,Vector b){
        return acos(Dot(a,b)/Length(a)/Length(b));
    }
    //逆时针旋转向量
    Vector Rotate(Vector a,double rad){
        return Vector(a.x*cos(rad)-a.y*sin(rad),a.x*sin(rad)+a.y*cos(rad));
    }
    //求p+v*t与q+w*t的交点,使用时确保Cross(v,w)不等于0
    Point GetlineIntersection(Point p,Vector v,Point q,Vector w){
        Vector u=p-q;
        double t=Cross(w,u)/Cross(v,w);
        return p+v*t;
    }
    //求p到直线ab的距离
    double DistanceToline(Point p,Point a,Point b){
        Vector v1=p-a,v2=b-a;
        return fabs(Cross(v1,v2)/Length(v2));
    }
    //求p到线段ab的距离
    double DistanceToSegment(Point p,Point a,Point b){
        if(a==b)return Length(p-a);
        Vector v1=b-a,v2=p-a,v3=p-b;
        if(dcmp(Dot(v1,v2)<0))return Length(p-a);
        else if(dcmp(Dot(v1,v3))>0)return Length(p-b);
        else return fabs(Cross(v1,v2)/Length(v1));
    }
    //线段a1a2与线段b1b2规范相交返回真
    bool SegmenProperIntersection(Point a1,Point a2,Point b1,Point b2){
        double c1=Cross(a2-a1,b1-a1),c2=Cross(a2-a1,b2-a1);
        double c3=Cross(b2-b1,a1-b1),c4=Cross(b2-b1,a2-b1);
        return dcmp(c1)*dcmp(c2)<0&&dcmp(c3)*dcmp(c4)<0;
    }
    //点p在线段a1a2上返回真
    bool OnSegment(Point p,Point a1,Point a2){
        return dcmp(Cross(a1-p,a2-p))==0&&dcmp(Dot(a1-p,a2-p))<0;
    }
    //点p在ab上的投影
    Point GetLineProjection(Point P,Point A,Point B)
    {
        Vector v=B-A;
        return A+v*(Dot(v,P-A)/Dot(v,v));
    }
    //与 x 轴的夹角,取值范围为 (-π,π]
    double angle(Vector v){
        return atan2(v.y,v.x);
    }
    //求线段a1,a2到线段b1,b2的最短距离
    double disSegmenttoSegment(Point a1,Point a2,Point b1,Point b2)
    {
        double ans=DistanceToSegment(a1,b1,b2);
        ans=min(ans,DistanceToSegment(a2,b1,b2));
        ans=min(ans,DistanceToSegment(b1,a1,a2));
        ans=min(ans,DistanceToSegment(b2,a1,a2));
        return ans;
    }
    

    线

    struct Line{
        Point p;//点
        Vector v;//向量
        double ang;
        Line(){}
        Line(Point p,Vector v):p(p),v(v){ang=atan2(v.y,v.x);}
        Point point(double t){return p+v*t;}
        bool operator <(const Line& L)const{
            return ang<L.ang;
        }
    };
    //点在直线的左边
    bool OnLeft(Line l,Point p)
    {
        return Cross(l.v,p-l.p)>0;
    }
    //两直线交点
    Point GetIntersection(Line a,Line b)
    {
        Vector u=a.p-b.p;
        double t=Cross(b.v,u)/Cross(a.v,b.v);
        return a.p+a.v*t;
    }
    

    多边形

    typedef vector<Point> Polygon;
    //多边形的有向面积,逆时针为正
    double PolygonArea(Polygon po) {
        int n = po.size();
        double area = 0.0;
        for(int i = 1; i < n-1; i++) {
            area += Cross(po[i]-po[0], po[i+1]-po[0]);
        }
        return area * 0.5;
    }
    //点是否在多边形内
    int isPointInPolygon(Point p,Point poly[],int n)
    {
        int wn=0;
        for(int i=0;i<n;i++){ 
            if(poly[i]==p||poly[(i+1)%n]==p||OnSegment(p,poly[i],poly[(i+1)%n]))return -1;
            int k=dcmp(Cross(poly[(i+1)%n]-poly[i],p-poly[i]));
            int d1=dcmp(poly[i].y-p.y);
            int d2=dcmp(poly[(i+1)%n].y-p.y);
            if(k>0&&d1<=0&&d2>0)
                wn++;
            if(k<0&&d2<=0&&d1>0)
                wn--;
        }
        if(wn!=0)return 1;//内部
        return 0;//外部
    }
    //求凸包,ch为返回凸包,m为凸包内点的数目,<=不允许点在边上,
    int ConvecHull(Point* p,int n,Point* ch)
    {
        sort(p,p+n);
        int m=0;
        for(int i=0;i<n;i++){
            while(m>1&&dcmp(Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0))m--;//注意<=与<的区别
            ch[m++]=p[i];
        }
        int k=m;
        for(int i=n-2;i>=0;i--){
            while(m>k&&dcmp(Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0))m--;//注意<=与<的区别
            ch[m++]=p[i];
        }
        if(n>1)m--;
        return m;
    }
    

    struct Circle{
        Point c;
        double r;
        Circle(){};
        Circle(Point c,double r=0):c(c),r(r){}
        Point point(double a){
            return Point(c.x+cos(a)*r,c.y+sin(a)*r);
        }
    };
    //求圆与直线的交点,t1,t2为(at+b)^2+(ct+d)^2=r^2的解,交点放入sol
    int getLineCircleIntersection(Line L,Circle C,double& t1,double& t2,vector<Point>& sol)
    {
        double a=L.v.x,b=L.p.x-C.c.x,c=L.v.y,d=L.p.y-C.c.y;
        double e=a*a+c*c,f=2*(a*b+c*d),g=b*b+d*d-C.r*C.r;
        double delta=f*f-4*e*g;
        if(dcmp(delta)<0)return 0;
        if(dcmp(delta)==0){
            t1=t2=-f/(2*e);
            sol.push_back(L.point(t1));
            return 1;
        }
        t1=(-f-sqrt(delta))/(2*e);
        sol.push_back(L.point(t1));
        t2=(-f+sqrt(delta))/(2*e);
        sol.push_back(L.point(t2));
        return 2;
    }
    //求圆与圆的交点,交点放入sol
    int getCircleCircleIntersection(Circle C1,Circle C2,vector<Point>& sol)
    {
        double d=Length(C1.c-C2.c);
        if(dcmp(d)==0){
            if(dcmp(C1.r-C2.r)==0)return -1;//两圆重合
            return 0;
        }
        if(dcmp(C1.r+C2.r-d)<0)return 0;
        if(dcmp(fabs(C1.r-C2.r)-d)>0)return 0;
        double  a=angle(C2.c-C1.c);//直线c1c2
        double da=acos((C1.r*C1.r+d*d-C2.r*C2.r)/(2*C1.r*d));//c1c2到c1p1的角
        Point p1=C1.point(a-da),p2=C1.point(a+da);
        sol.push_back(p1);
        if(p1==p2)return 1;
        sol.push_back(p2);
        return 2;
    }
    //求直线与圆的切线
    int getTangents(Point p,Circle C,Vector* v){
        Vector u=C.c-p;
        double dist=Length(u);
        if(dcmp(dist-C.r)<0)return 0;
        else if(dcmp(dist-C.r)==0){
            v[0]=Rotate(u,PI/2);
            return 1;
        }
        else {
            double ang=asin(C.r/dist);
            v[0]=Rotate(u,-ang);
            v[1]=Rotate(u,+ang);
            return 2;
        }
    }
    

    半平面交

    int HalfplaneIntersection(Line* L,int n,Point* poly)
    {
        sort(L,L+n);
        int first,last;
        Point *p=new Point[n];//会在函数内开大容量数组,请在使用时注意开成全局!!!!!
        Line *q=new Line[n];//会在函数内开大容量数组,请在使用时注意开成全局!!!!!
        q[first=last=0]=L[0];
        for(int i=1;i<n;i++){
            while(first<last&&!OnLeft(L[i],p[last-1]))last--;
            while(first<last&&!OnLeft(L[i],p[first]))first++;
            q[++last]=L[i];
            if(fabs(Cross(q[last].v,q[last-1].v))<eps){
                last--;
                if(OnLeft(q[last],L[i].p))q[last]=L[i];
            }
            if(first<last)p[last-1]=GetIntersection(q[last-1],q[last]);
        }
        while(first<last&&!OnLeft(q[first],p[last-1]))last--;
        if(last-first<=1)return 0;
        p[last]=GetIntersection(q[last],q[first]);
    
        int m=0;
        for(int i=first;i<=last;i++)poly[m++]=p[i];
        return m;
    }//会在函数内开大容量数组,请在使用时注意开成全局
    

    平面直线图(PSGL)

    struct Edge
    {
        int from, to; // 起点,终点,左边的面编号
        double ang;
        Edge(int f,int t,double a):from(f),to(t),ang(a) {}
    };
    // 平面直线图(PSGL)实现
    struct PSLG {
        int n, m, face_cnt;
        double x[maxn], y[maxn];
        vector<Edge> edges;
        vector<int> G[maxn];
        int vis[maxn*2];  // 每条边是否已经访问过
        int left[maxn*2]; // 左面的编号(该边在哪个面内)
        int prev[maxn*2]; // 相同起点的上一条边(即顺时针旋转碰到的下一条边)的编号
    
        vector<Polygon> faces;
        double area[maxn]; // 每个polygon的面积
    
        void init(int n) {
            this->n = n;
            for(int i = 0; i < n; i++) G[i].clear();
            edges.clear();
            faces.clear();
        }
    
        // 有向线段from->to的极角
        double getAngle(int from, int to) {
            return atan2(y[to]-y[from], x[to]-x[from]);
        }
    
        void AddEdge(int from, int to) {
            edges.push_back((Edge){from, to, getAngle(from, to)});
            edges.push_back((Edge){to, from, getAngle(to, from)});
            m = edges.size();
            G[from].push_back(m-2);
            G[to].push_back(m-1);
        }
    
        // 找出faces并计算面积
        void Build() {
            for(int u = 0; u < n; u++) {
                // 给从u出发的各条边按极角排序
                int d = G[u].size();
                for(int i = 0; i < d; i++)
                    for(int j = i+1; j < d; j++) // 这里偷个懒,假设从每个点出发的线段不会太多
                        if(edges[G[u][i]].ang > edges[G[u][j]].ang) swap(G[u][i], G[u][j]);
                for(int i = 0; i < d; i++)
                    prev[G[u][(i+1)%d]] = G[u][i];  //u点出发的第i条边顺时针转的第一条边是prev[i]
            }
    
            memset(vis, 0, sizeof(vis));
            face_cnt = 0;
            for(int u = 0; u < n; u++)
                for(int i = 0; i < G[u].size(); i++) {
                    int e = G[u][i];  //逆时针转的第i条边
                    if(!vis[e]) { // 逆时针找圈
                        face_cnt++;
                        Polygon poly;
                        for(;;) {
                            vis[e] = 1; left[e] = face_cnt;
                            int from = edges[e].from;
                            poly.push_back(Point(x[from], y[from]));
                            //cout<<x[from]<<"   "<<y[from]<<"   ";
                            e = prev[e^1];                    //反向边顺时针第一条
                            if(e == G[u][i]) break;       //回到原点
                            assert(vis[e] == 0);
                        }
                        //cout<<endl;
                        faces.push_back(poly);
                    }
                }
    
            for(int i = 0; i < faces.size(); i++) {
                area[i] = PolygonArea(faces[i]);
            }
        }
    };
    

    旋转卡壳

    //向量(b-a)与向量(c-a)的叉积,相当于三角形abc的有向面积的2倍
    double cross(Point a,Point b,Point c)
    {
        return Cross(b-a,c-a);
    }
    
    //求凸包内最远点
    long long getmaxdistance(Point a[],int n)
    {
        int j=2;
        double ans=0;
        a[n]=a[0];//方便写下一个点,避免取模
        for(int i=0;i<n;i++){
            while(fabs(cross(a[i],a[i+1],a[j+1]))>fabs(cross(a[i],a[i+1],a[j])))j=(j+1)%n;
            //通过比较面积大小,比较到直线的距离
            ans=max(ans,Length2(a[j]-a[i]));
        }
        return ans;
    }
    
    //求2个凸包间的最短距离
    double getmindistance(Point p1[],Point p2[],int n1,int n2)
    {
        int i=0,j=0;
        for(int k=0;k<n1;k++){
            if(p1[k].y<p1[i].y)i=k;//找出p1中的y最小值的点
        }
        for(int k=0;k<n2;k++){
            if(p2[k].y>p2[j].y)j=k;//找出p2中的y最大值的点
        }
        p1[n1]=p1[0];
        p2[n2]=p2[0];
        double ans=99999999999;
        for(int k=0;k<n1;k++){
            //循环n1次,相当于求p1中每一条边与p2的最近距离
            while((cross(p1[i],p1[i+1],p2[j+1])-cross(p1[i],p1[i+1],p2[j]))>eps)
                j=(j+1)%n2;
            ans=min(ans,disSegmenttoSegment(p1[i],p1[i+1],p2[j],p2[j+1]));//求线段间的最短距离
            i=(i+1)%n1;
        }
        return ans;
    }
    
    //求凸包的内4个点组成的最大四边形面积
    double solve(Point a[],int n)
    {
        a[n]=a[0];
        int p1,p2;
        double ans=0;
        for(int i=0;i<n;i++){
            p1=(i+0)%n;
            p2=(i+1)%n;
            for(int j=i+1;j<n;j++){
                while(cross(a[i],a[j],a[p1+1])<cross(a[i],a[j],a[p1]))p1=(p1+1)%n;
                while(cross(a[i],a[j],a[p2+1])>cross(a[i],a[j],a[p2]))p2=(p2+1)%n;
                ans=max(ans,cross(a[i],a[j],a[p2])-cross(a[i],a[j],a[p1]));
            }
        }
        return ans;
    }
    

    三维几何

    基础点面

    struct Point3{
        double x,y,z;
        Point3(double x=0,double y=0,double z=0):x(x),y(y),z(z){}
    };
    
    typedef Point3 Vector3;
    
    Vector3 operator +(Vector3 A,Vector3 B){
        return Vector3(A.x+B.x,A.y+B.y,A.z+B.z);
    }
    Vector3 operator -(Vector3 A,Vector3 B){
        return Vector3(A.x-B.x,A.y-B.y,A.z-B.z);
    }
    Vector3 operator *(Vector3 A,double p){
        return Vector3(A.x*p,A.y*p,A.z*p);
    }
    Vector3 operator /(Vector3 A,double p){
        return Vector3(A.x/p,A.y/p,A.z/p);
    }
    int dcmp(double a){
        if(fabs(a)<eps)return 0;
        else if(a>0)return 1;
        return -1;
    }
    
    bool operator ==(Vector3 a,Vector3 b){
        return dcmp(a.x-b.x)==0&&dcmp(a.y-b.y)==0&&dcmp(a.z-b.z)==0;
    }
    double Dot(Vector3 A,Vector3 B){
        return A.x*B.x+A.y*B.y+A.z*B.z;
    }
    double Length(Vector3 A){
        return sqrt(Dot(A,A));
    }
    double Angle(Vector3 A,Vector3 B){
        return acos(Dot(A,B/Length(A)/Length(B)));
    }
    
    //点p到平面p0-n的距离。n必须为单位向量
    double DistancetoPlane(const Point3& p,const Point3& p0,const Vector3& n){
        return fabs(Dot(p-p0,n));
    }
    
    //点p在平面p0-n上的投影。n必须为单位向量
    Point3 GetPlaneProjection(const Point3& p,const Point3& p0,const Vector3& n){
        return p-n*Dot(p-p0,n);
    }
    
    //直线p1-p2到平面p0-n的交点。假设交点唯一存在
    Point3 LinePlaneIntersection(Point3 p1,Point3 p2,Point3 p0,Vector3 n){
        Vector3 v=p2-p1;
        double t=(Dot(n,p0-p1))/Dot(n,p2-p1);
        return p1+v*t;
    }
    
    Vector3 Cross(Vector3 A,Vector3 B){
        return Vector3(A.y*B.z-A.z*B.y,A.z*B.x-A.x*B.z,A.x*B.y-A.y*B.x);
    }
    
    double Area2(Point3 A,Point3 B,Point3 C){
        return Length(Cross(B-A,C-A));
    }
    
    //点p在△p0p1p2中
    bool PointInTri(Point3 p,Point3 p0,Point3 p1,Point3 p2){
        double area1=Area2(p,p0,p1);
        double area2=Area2(p,p1,p2);
        double area3=Area2(p,p2,p0);
        return dcmp(area1+area2+area3-Area2(p0,p1,p2))==0;
    }
    
    //△p0p1p2是否和线段ab相交
    bool TriSegIntersection(Point3 p0,Point3 p1,Point3 p2,Point3 a,Point3 b,Point3& p)
    {
        Vector3 n=Cross(p1-p0,p2-p0);
        if(dcmp(Dot(n,b-a))==0)return false;
        else {
            double t=Dot(n,p0-a)/Dot(n,b-a);
            if(dcmp(t)<0||dcmp(t-1)>0)return false;
            p=a+(b-a)*t;
            return PointInTri(p,p0,p1,p2);
        }
    }
    
    //点p到直线ab的距离
    double DistancetoLine(Point3 p,Point3 a,Point3 b)
    {
        Vector3 v1=b-a,v2=p-a;
        return Length(Cross(v1,v2)/Length(v1));
    }
    
    //点p到线段ab的距离
    double DistanceToSegment(Point3 p,Point3 a,Point3 b){
        if(a==b)return Length(p-a);
        Vector3 v1=b-a,v2=p-a,v3=p-b;
        if(dcmp(Dot(v1,v2)<0))return Length(v2);
        else if(dcmp(Dot(v1,v3))>0)return Length(v3);
        else return Length(Cross(v1,v2))/Length(v1);
    }
    
    //返回ab,ac,ad的混合积。它也等于四面体的有向体积的6倍
    double Volume6(Point3 a,Point3 b,Point3 c,Point3 d)
    {
        return Dot(d-a,Cross(b-a,c-a));
    }
    

    凸包

    struct Face{
        int v[3];
        Vector3 normal(Point3 *p)const{
            return Cross(p[v[1]]-p[v[0]],p[v[2]]-p[v[0]]);
        }
        int cansee(Point3 *p,int i)const{
            return Dot(p[i]-p[v[0]],normal(p))>0?1:0;
        }
    };
    const int N=1e3;
    int vis[N][N];
    //三维凸包,记得使用前扰动点,以避免特殊情况
    vector<Face> CH3D(Point3 *p,int n)
    {
        vector<Face> cur;
        //记得提前扰动
        cur.push_back((Face){{0,1,2}});
        cur.push_back((Face){{2,1,0}});
        for(int i=3;i<n;i++){
            vector<Face> next;
            for(int j=0;j<cur.size();j++){
                Face& f=cur[j];
                int res=f.cansee(p,i);
                if(!res)next.push_back(f);
                for(int k=0;k<3;k++)vis[f.v[k]][f.v[(k+1)%3]]=res;
            }
            for(int j=0;j<cur.size();j++){
                for(int k=0;k<3;k++){
                    int a=cur[j].v[k],b=cur[j].v[(k+1)%3];
                    if(vis[a][b]!=vis[b][a]&&vis[a][b]){
                        next.push_back((Face){{a,b,i}});
                    }
                }
            }
            cur=next;
        }
        return cur;
    }
    double rand01(){
        return rand()/(double)RAND_MAX;
    }
    double randeps(){
        return (rand01()-0.5)*eps;
    }
    Point3 add_noise(Point3 p)
    {
        return Point3(p.x+randeps(),p.y+randeps(),p.z+randeps());
    }
    
    
    我现在最大的问题就是人蠢而且还懒的一批。
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  • 原文地址:https://www.cnblogs.com/pot-a-to/p/11140086.html
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