[计算几何]【学习笔记】
学计算几何专题已经两年了(还不是因为刚刚过年了.....)
哼 都写完了网页崩溃全没了 生气了 哼 本来已经12点了超想睡觉的现在又要重写一遍 哼 我不睡觉了
先扔一个完整版方便复制
//By Candy? #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> using namespace std; typedef long long ll; const int N=1e4,INF=1e9; inline int read(){ char c=getchar();int x=0,f=1; while(c<'0'||c>'9'){if(c=='-')f=-1; c=getchar();} while(c>='0'&&c<='9'){x=x*10+c-'0'; c=getchar();} return x*f; } const double eps=1e-8; const double pi=acos(-1); inline int sgn(double x){ if(abs(x)<eps) return 0; else return x<0?-1:1; } struct Vector{ double x,y; Vector(double a=0,double b=0):x(a),y(b){} bool operator <(const Vector &a)const{ return sgn(x-a.x)<0||(sgn(x-a.x)==0&&sgn(y-a.y)<0); } void print(){printf("%lf %lf\n",x,y);} }; typedef Vector Point; Vector operator +(Vector a,Vector b){return Vector(a.x+b.x,a.y+b.y);} Vector operator -(Vector a,Vector b){return Vector(a.x-b.x,a.y-b.y);} Vector operator *(Vector a,double b){return Vector(a.x*b,a.y*b);} Vector operator /(Vector a,double b){return Vector(a.x/b,a.y/b);} bool operator ==(Vector a,Vector b){return sgn(a.x-b.x)==0&&sgn(a.y-b.y)==0;} double Dot(Vector a,Vector b){return a.x*b.x+a.y*b.y;} double Cross(Vector a,Vector b){return a.x*b.y-a.y*b.x;} double Len(Vector a){return sqrt(Dot(a,a));} double Len2(Vector a){return Dot(a,a);} double Angle(Vector a,Vector b){ return acos(Dot(a,b)/Len(a)/Len(b)); } Vector Normal(Vector a){ return Vector(-a.y,a.x);//counterClockwise } Vector Rotate(Vector a,double rad){ return Vector(a.x*cos(rad)-a.y*sin(rad),a.x*sin(rad)+a.y*cos(rad)); } //Line:use p and v //struct Line{ // Point p; // Vector v; // Line(){} // Line(Point p,Vector v):p(p),v(v){} // bool operator <(const Line a)const{ // return sgn(Cross(v,a.v))>=0; // } //}; //bool OnLeft(Line l,Point p){ // return sgn(Cross(l.v,p-l.p))>=0; //} //Point LI(Line a,Line b){ // Vector v=a.p-b.p; // double t=Cross(b.v,v)/Cross(a.v,b.v); // return a.p+a.v*t; //} //Line:use s and t struct Line{ Point s,t; Line(){} Line(Point a,Point b):s(a),t(b){} bool operator <(Line a)const{ return sgn(Cross(t-s,a.t-a.s))>=0; } }; double DisTL(Point p,Point a,Point b){ Vector v1=b-a,v2=p-a; return abs(Cross(v1,v2)/Len(v1)); } double DisTS(Point p,Point a,Point b){ if(a==b) return Len(p-a); Vector v1=b-a,v2=p-a,v3=p-b; if(sgn(Dot(v1,v2))<0) return Len(v2); else if(sgn(Dot(v1,v3))>0) return Len(v3); else return abs(Cross(v1,v2)/Len(v1)); } bool OnLeft(Line l,Point p){ return sgn(Cross(l.t-l.s,p-l.s))>=0; } bool OnSeg(Point p,Point a,Point b){ return DisTL(p,a,b)==0&&sgn(Dot(p-a,p-b)<0&&!(p==a))&&!(p==b); } Point LI(Line a,Line b){ Vector v=a.s-b.s,v1=a.t-a.s,v2=b.t-b.s; double t=Cross(v2,v)/Cross(v1,v2); return a.s+v1*t; } bool isLSI(Line l1,Line l2){ Vector v=l1.t-l1.s,u=l2.s-l1.s,w=l2.t-l1.s; return sgn(Cross(v,u))!=sgn(Cross(v,w)); } bool isSSI(Line l1,Line l2){ return isLSI(l1,l2)&&isLSI(l1,l2); } //---chong he //bool isSSI(Line l1,Line l2){ // Vector v1=l1.t-l1.s,v2=l2.t-l2.s; // if(sgn(Cross(v1,v2))==0){ // int flag=0; // Vector u=l2.s-l1.s,w=l2.t-l1.s; // if(sgn(Dot(u,w))<0) flag=1; // u=l2.s-l1.t,w=l2.t-l1.t; // if(sgn(Dot(u,w))<0) flag=1; // return flag; // } // else return isLSI(l1,l2)&&isLSI(l2,l1); //} Point Circumcenter(Point a,Point b,Point c){ Point p=(a+b)/2,q=(a+c)/2; Vector v=Normal(b-a),u=Normal(c-a); if(sgn(Cross(v,u))==0){ if(sgn(Len(a-b)+Len(b-c)-Len(a-c))==0) return (a+c)/2; if(sgn(Len(a-b)+Len(a-c)-Len(b-c))==0) return (b+c)/2; if(sgn(Len(a-c)+Len(b-c)-Len(a-b))==0) return (a+b)/2; } return LI(Line(p,p+v),Line(q,q+u)); } Point Barycenter(Point a,Point b,Point c){ return (a+b+c)/3; } bool cmpPolar(Point a,Point b){ return sgn(Cross(a,b))>0; } int PointInPolygon(Point p,Point poly[],int n){ int wn=0; for(int i=1;i<=n;i++){ if(sgn(DisTS(p,poly[i],poly[i%n+1]))==0) return -1; int k=sgn(Cross(poly[i%n+1]-poly[i],p-poly[i])), d1=sgn(poly[i].y-p.y),d2=sgn(poly[i%n+1].y-p.y); if(k>0&&d1<=0&&d2>0) wn++; if(k<0&&d2<=0&&d1>0) wn--; } return (bool)wn; } double PolygonArea(Point p[],int n){ double s=0; for(int i=2;i<n;i++) s+=Cross(p[i]-p[1],p[i+1]-p[1]); return abs(s/2); } bool isConvex(Point poly[],int n){ int last=0,now=0; for(int i=1;i<=n;i++){ now=sgn(Cross(poly[i%n+1]-poly[i],poly[(i+1)%n+1]-poly[i%n+1])); if(last==0||now==0||now*last>0) last=now; else return false; } return true; } int ConvexHull(Point p[],int n,Point ch[]){//cannot handle repeat point sort(p+1,p+1+n); int m=0; for(int i=1;i<=n;i++){ while(m>1&&sgn(Cross(ch[m]-ch[m-1],p[i]-ch[m-1]))<=0) m--; ch[++m]=p[i]; } int k=m; for(int i=n-1;i>=1;i--){//n-1 while(m>k&&sgn(Cross(ch[m]-ch[m-1],p[i]-ch[m-1]))<=0) m--; ch[++m]=p[i]; } if(n>1) m--;//the first rpoint return m; } double RotatingCalipers(Point p[],int n){ if(n==1) return 0; if(n==2) return Len(p[1]-p[2]); int now=1; double ans=0; p[n+1]=p[1]; for(int i=1;i<=n;i++){ while(sgn(DisTL(p[now],p[i],p[i+1])-DisTL(p[now+1],p[i],p[i+1]))<=0) now=now%n+1; ans=max(ans,Len(p[now]-p[i])); ans=max(ans,Len(p[now]-p[i+1])); } return ans; } void iniPolygon(Point p[],int &n,double inf){ n=0; p[++n]=Point(-inf,-inf); p[++n]=Point(inf,-inf); p[++n]=Point(inf,inf); p[++n]=Point(-inf,inf); } Point t[N];int tn; void CutPolygon(Point p[],int &n,Point a,Point b){//get the left of a->b tn=0; Point c,d,e; for(int i=1;i<=n;i++){ c=p[i],d=p[i%n+1]; if(sgn(Cross(b-a,c-a))>=0) t[++tn]=c; if(isLSI(Line(a,b),Line(c,d))){ e=LI(Line(a,b),Line(c,d)); t[++tn]=e; } } n=tn;for(int i=1;i<=n;i++)p[i]=t[i]; } // iniPolygon(q,m,INF); // for(int i=1;i<=n;i++) CutPolygon(q,m,p[i%n+1],p[i]); double minCircleCover(Point p[],int n,Point &c){ random_shuffle(p+1,p+1+n); c=p[1]; double r=0; for(int i=2;i<=n;i++) if(sgn(Len(c-p[i])-r)>0){ c=p[i],r=0; for(int j=1;j<i;j++) if(sgn(Len(c-p[j])-r)>0){ c=(p[i]+p[j])/2,r=Len(c-p[i]); for(int k=1;k<j;k++) if(sgn(Len(c-p[k])-r)>0){ c=Circumcenter(p[i],p[j],p[k]); r=Len(c-p[i]); } } } return r; } //jie xi ji he double a,b,c; double f(Point p){return a*p.x+b*p.y+c;} Point abcLI(Line l){ double u=abs(f(l.s)),v=abs(f(l.t)); return Point(l.s.x*v+l.t.x*u,l.s.y*v+l.t.y*u)/(u+v); } double F(double x){return x;} //function inline double cal(double l,double r,double fl,double fr,double fm){ return (fl+fr+4*fm)*(r-l)/6; } double Simpson(double l,double r,double now,double fl,double fr,double fm){ double mid=(l+r)/2,flm=F((l+mid)/2),frm=F((mid+r)/2); double p=cal(l,mid,fl,fm,flm),q=cal(mid,r,fm,fr,frm); if(sgn(now-p-q)==0) return now; else return Simpson(l,mid,p,fl,fm,flm)+Simpson(mid,r,q,fm,fr,frm); } //sao miao xian
并不想多说什么 ljw的课件已经很好了
1.精度问题
const double eps=1e-8; inline int sgn(double x){ if(abs(x)<eps) return 0; else return x<0?-1:1; }
2.向量 点
向量+-*／ 比较点积 cosΘ
叉积 sinΘ aXb>0 b在a的左面(逆时针方向)
double Dot(Vector a,Vector b){return a.x*b.x+a.y*b.y;} double Cross(Vector a,Vector b){return a.x*b.y-a.y*b.x;} double Len(Vector a){return sqrt(Dot(a,a));} double Len2(Vector a){return Dot(a,a);} double Angle(Vector a,Vector b){ return acos(Dot(a,b)/Len(a)/Len(b)); }
法向量 (-y,x) 逆时针90
旋转 想一个与坐标轴平行的直角然后一推就可以了
Vector Normal(Vector a){ return Vector(-a.y,a.x);//counterClockwise } Vector Rotate(Vector a,double rad){ return Vector(a.x*cos(rad)-a.y*sin(rad),a.x*sin(rad)+a.y*cos(rad)); }
3.直线 线段
感觉用两点比较方便
struct Line{ Point s,t; Line(){} Line(Point a,Point b):s(a),t(b){} bool operator <(Line a)const{ return sgn(Cross(t-s,a.t-a.s))>=0; } };
点到直线距离：面积法
线段最短距离：用点积特判是点到端点到距离的情况
点线位置关系 点在线段上
double DisTL(Point p,Point a,Point b){ Vector v1=b-a,v2=p-a; return abs(Cross(v1,v2)/Len(v1)); } double DisTS(Point p,Point a,Point b){ if(a==b) return Len(p-a); Vector v1=b-a,v2=p-a,v3=p-b; if(sgn(Dot(v1,v2))<0) return Len(v2); else if(sgn(Dot(v1,v3))>0) return Len(v3); else return abs(Cross(v1,v2)/Len(v1)); } bool OnLeft(Line l,Point p){ return sgn(Cross(l.t-l.s,p-l.s))>=0; } bool OnSeg(Point p,Point a,Point b){ return DisTL(p,a,b)==0&&sgn(Dot(p-a,p-b)<0&&!(p==a))&&!(p==b); }
直线相交交点：面积求比例，正负画个图看一下是否符合就行了
直线与线段相交：端点叉积不同
线段与线段相交：两遍直线与线段(有时候特判重合)
Point LI(Line a,Line b){ Vector v=a.s-b.s,v1=a.t-a.s,v2=b.t-b.s; double t=Cross(v2,v)/Cross(v1,v2); return a.s+v1*t; } bool isLSI(Line l1,Line l2){ Vector v=l1.t-l1.s,u=l2.s-l1.s,w=l2.t-l1.s; return sgn(Cross(v,u))!=sgn(Cross(v,w)); } bool isSSI(Line l1,Line l2){ return isLSI(l1,l2)&&isLSI(l1,l2); }
4.三角形
外心：两个中垂线交点，特判三点共线
重心：三个点平均
Point Circumcenter(Point a,Point b,Point c){ Point p=(a+b)/2,q=(a+c)/2; Vector v=Normal(b-a),u=Normal(c-a); if(sgn(Cross(v,u))==0){ if(sgn(Len(a-b)+Len(b-c)-Len(a-c))==0) return (a+c)/2; if(sgn(Len(a-b)+Len(a-c)-Len(b-c))==0) return (b+c)/2; if(sgn(Len(a-c)+Len(b-c)-Len(a-b))==0) return (a+b)/2; } return LI(Line(p,p+v),Line(q,q+u)); } Point Barycenter(Point a,Point b,Point c){ return (a+b+c)/3; }
5.多边形
bool cmpPolar(Point a,Point b){ return sgn(Cross(a,b))>0; }
double PolygonArea(Point p[],int n){ double s=0; for(int i=2;i<n;i++) s+=Cross(p[i]-p[1],p[i+1]-p[1]); return abs(s/2); }
选一个基准点，不停叉积
判断点在多边形内：射线法
- 引一条向右的水平射线
- 注意判断点在多边形上的情况
- 逆时针穿过边界，转数+1，顺时针则-1
- 注意射线穿过顶点的情况 下算上不算
int PointInPolygon(Point p,Point poly[],int n){ int wn=0; for(int i=1;i<=n;i++){ if(sgn(DisTS(p,poly[i],poly[i%n+1]))==0) return -1; int k=sgn(Cross(poly[i%n+1]-poly[i],p-poly[i])), d1=sgn(poly[i].y-p.y),d2=sgn(poly[i%n+1].y-p.y); if(k>0&&d1<=0&&d2>0) wn++; if(k<0&&d2<=0&&d1>0) wn--; } return (bool)wn; }
凸包：
判断凸包：相邻边叉积同号
求凸包：Graham扫描法 我不重写了！细节见代码
bool isConvex(Point poly[],int n){ int last=0,now=0; for(int i=1;i<=n;i++){ now=sgn(Cross(poly[i%n+1]-poly[i],poly[(i+1)%n+1]-poly[i%n+1])); if(last==0||now==0||now*last>0) last=now; else return false; } return true; } int ConvexHull(Point p[],int n,Point ch[]){//cannot handle repeat point sort(p+1,p+1+n); int m=0; for(int i=1;i<=n;i++){ while(m>1&&sgn(Cross(ch[m]-ch[m-1],p[i]-ch[m-1]))<=0) m--; ch[++m]=p[i]; } int k=m; for(int i=n-1;i>=1;i--){//n-1 while(m>k&&sgn(Cross(ch[m]-ch[m-1],p[i]-ch[m-1]))<=0) m--; ch[++m]=p[i]; } if(n>1) m--;//the first rpoint return m; }
旋转卡壳
- 被凸包上被一对平行直线卡住的点对叫对踵点
- 答案一定出在一对对踵点上
- 尝试用通过旋转一对平行直线，枚举所有对踵点
- 按逆时针顺序枚举凸包上一条边，则凸包上到该边所在直线最远的点也单调逆时针旋转，相当于用一条直线卡对面一个顶点
double RotatingCalipers(Point p[],int n){ if(n==1) return 0; if(n==2) return Len(p[1]-p[2]); int now=1; double ans=0; p[n+1]=p[1]; for(int i=1;i<=n;i++){ while(sgn(DisTL(p[now],p[i],p[i+1])-DisTL(p[now+1],p[i],p[i+1]))<=0) now=now%n+1; ans=max(ans,Len(p[now]-p[i])); ans=max(ans,Len(p[now]-p[i+1])); } return ans; }
注意特判；变种很多 见题目
半平面交
- 初始化时加上一个范围巨大的“框”
- 每次拿一个新的半平面切割原先的凸集
- 保留在新加直线左边的点，删除右边的，有向直线与多边形相交产生的新的点加入到新多边形内
void iniPolygon(Point p[],int &n,double inf){ n=0; p[++n]=Point(-inf,-inf); p[++n]=Point(inf,-inf); p[++n]=Point(inf,inf); p[++n]=Point(-inf,inf); } Point t[N];int tn; void CutPolygon(Point p[],int &n,Point a,Point b){//get the left of a->b tn=0; Point c,d,e; for(int i=1;i<=n;i++){ c=p[i],d=p[i%n+1]; if(sgn(Cross(b-a,c-a))>=0) t[++tn]=c; if(isLSI(Line(a,b),Line(c,d))){ e=LI(Line(a,b),Line(c,d)); t[++tn]=e; } } n=tn;for(int i=1;i<=n;i++)p[i]=t[i]; }
最小圆覆盖：三点确定圆 随即增量法 O(n)
double minCircleCover(Point p[],int n,Point &c){ random_shuffle(p+1,p+1+n); c=p[1]; double r=0; for(int i=2;i<=n;i++) if(sgn(Len(c-p[i])-r)>0){ c=p[i],r=0; for(int j=1;j<i;j++) if(sgn(Len(c-p[j])-r)>0){ c=(p[i]+p[j])/2,r=Len(c-p[i]); for(int k=1;k<j;k++) if(sgn(Len(c-p[k])-r)>0){ c=Circumcenter(p[i],p[j],p[k]); r=Len(c-p[i]); } } } return r; }
6.辛普森积分
用二次函数拟合
我也是看过托马斯微积分的....
double F(double x){return x;} //function inline double cal(double l,double r,double fl,double fr,double fm){ return (fl+fr+4*fm)*(r-l)/6; } double Simpson(double l,double r,double now,double fl,double fr,double fm){ double mid=(l+r)/2,flm=F((l+mid)/2),frm=F((mid+r)/2); double p=cal(l,mid,fl,fm,flm),q=cal(mid,r,fm,fr,frm); if(sgn(now-p-q)==0) return now; else return Simpson(l,mid,p,fl,fm,flm)+Simpson(mid,r,q,fm,fr,frm); }
7.解析几何
求直线与直线方程交点的神奇方法
double a,b,c; double f(Point p){return a*p.x+b*p.y+c;} Point abcLI(Line l){ double u=abs(f(l.s)),v=abs(f(l.t)); return Point(l.s.x*v+l.t.x*u,l.s.y*v+l.t.y*u)/(u+v); }
8.扫描线：见题目
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