计算几何模板

既然咱负责计算几何，就多刷一些题吧ww

模板改编自刘汝佳

点，线基础部分：

const double pi=acos(-1.0);
int dcmp(double x) {if(fabs(x) < eps) return 0; else return x < 0 ? -1 : 1; }
struct Vector
{
double x, y;
Vector (double x=0, double y=0) :x(x),y(y) {}
Vector operator + (const Vector &B) const { return Vector (x+B.x,y+B.y); }
Vector operator - (const Vector &B) const { return Vector(x - B.x, y - B.y); }
Vector operator * (const double &p) const { return Vector(x*p, y*p); }
Vector operator / (const double &p) const { return Vector(x/p, y/p); }
double operator * (const Vector &B) const { return x*B.x + y*B.y;}//点积
double operator ^ (const Vector &B) const { return x*B.y - y*B.x;}//叉积
bool operator < (const Vector &b) const { return x < b.x || (x == b.x && y < b.y); }
bool operator ==(const Vector &b) const { return dcmp(x-b.x) == 0 && dcmp(y-b.y) == 0; }
};
typedef Vector Point;
Point Read(){double x, y;scanf("%lf%lf", &x, &y);return Point(x, y);}
double Length(Vector A){ return sqrt(A*A); }//向量的模
double Angle(Vector A, Vector B){return acos(A*B / Length(A) / Length(B)); }//向量的夹角，返回值为弧度
double Area2(Point A, Point B, Point C){ return (B-A)^(C-A); }//向量AB叉乘AC的有向面积
Vector VRotate(Vector A, double rad){return Vector(A.x*cos(rad) - A.y*sin(rad), A.x*sin(rad) + A.y*cos(rad));}//向量A旋转rad弧度
Point  PRotate(Point A, Point B, double rad){return A + VRotate(B-A, rad);}//将B点绕A点旋转rad弧度
Vector Normal(Vector A){double l = Length(A);return Vector(-A.y/l, A.x/l);}//求向量A向左旋转90°的单位法向量,调用前确保A不是零向量

Point GetLineIntersection/*求直线交点，调用前要确保两条直线有唯一交点*/(Point P, Vector v, Point Q, Vector w){double t = (w^(P - Q)) / (v^w);return P + v*t;}//在精度要求极高的情况下，可以自定义分数类
double DistanceToLine/*P点到直线AB的距离*/(Point P, Point A, Point B){Vector v1 = B - A, v2 = P - A;return fabs(v1^v2) / Length(v1);}//不加绝对值是有向距离
double DistanceToSegment/*点到线段的距离*/(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(v1*v2)<0) return Length(v2);else
if (dcmp(v1*v3)>0) return Length(v3);else
return fabs(v1^v2)/Length(v1);
}

Point GetLineProjection/*点在直线上的射影*/(Point P, Point A, Point B)
{
Vector v=B-A;
return A+v*((v*(P-A))/(v*v));
}

bool OnSegment/*判断点是否在线段上(含端点)*/(Point P,Point a1,Point a2)
{
Vector v1=a1-P,v2=a2-P;
if (dcmp(v1^v2)==0 && min(a1.x,a2.x)<=P.x  && P.x<=max(a1.x,a2.x)  && min(a1.y,a2.y)<=P.y && P.y<=max(a1.y,a2.y)) return true;
return false;
}

bool SegmentInter/*线段相交判定*/(Point a1, Point a2, Point b1, Point b2)
{
//if (OnSegment(a1,b1,b2) || OnSegment(a2,b1,b2) || OnSegment(b1,a1,a2) || OnSegment(b2,a1,a2)) return 1;
//如果只判断线段规范相交（不算交点）,上面那句可以删掉
double c1=(a2-a1)^(b1-a1),c2=(a2-a1)^(b2-a1);
double c3=(b2-b1)^(a1-b1),c4=(b2-b1)^(a2-b1);
return dcmp(c1)*dcmp(c2)<0 && dcmp(c3)*dcmp(c4)<0;
}

bool InTri/*判断点是否在三角形内*/(Point P, Point a,Point b,Point c)
{
if (dcmp(fabs((c-a)^(c-b))-fabs((P-a)^(P-b))-fabs((P-b)^(P-c))-fabs((P-a)^(P-c)))==0) return true;
return false;
}

double PolygonArea/*求多边形面积，注意凸包P序号从0开始*/(Point *P ,int n)
{
double ans = 0.0;
for(int i=1;i<n-1;i++)
ans+=(P[i]-P[0])^(P[i+1]-P[0]);
return ans/2;
}
bool CrossOfSegAndLine/*判断线段是否与直线相交*/(Point a1,Point a2,Point b1,Vector b2)
{
if (OnSegment(b1,a1,a2) || OnSegment(b1+b2,a1,a2)) return true;
return dcmp(b2^(a1-b1))*dcmp(b2^(a2-b1))<0;
}

半平面交

struct Line//有向直线
{
Point p;
Vector v;
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
{
if (fabs(ang-L.ang)<eps)
{
return ((v)^(L.p-p))<0;
}
return ang < L.ang;
}
};

bool OnRight(Line L,Point p)//判断是否p点在有向直线右边
{
return ((L.v^(p-L.p))<0);
}
Point GetIntersection (Line a,Line b)//求两个有向直线的交点
{
Vector u=a.p-b.p;
double t=(b.v^u)/(a.v^b.v);
return a.p+a.v*t;
}

bool IsParallel(Line a,Line b)//判断两条有向直线是否平行
{
if (dcmp(a.v ^ b.v)==0) return 1;
return 0;
}

Line q[MAXN];
int HalfPlane(Line *L,int n,Point *poly)//半平面交，输入直线数组L(需要保证逆时针顺序)，n是直线的个数，poly用于输出，返回值为poly中点的个数
{
sort(L,L+n);
int m,i;
for (m=i=1;i<n;i++)
{
if (dcmp(L[i].ang-L[i-1].ang)!=0) L[m++]=L[i];
}
n=m;
int l=0,r=1;
q[0]=L[0];q[1]=L[1];
for (int i=2;i<n;i++)
{
//if (IsParallel(q[r],q[r-1]) || IsParallel(q[l],q[l+1])) return 0;
while (l < r && OnRight(L[i],GetIntersection(q[r-1],q[r]))) r--;
while (l < r && OnRight(L[i],GetIntersection(q[l],q[l+1]))) l++;
q[++r]=L[i];
}
while (l<r && OnRight(q[l],GetIntersection(q[r-1],q[r]))) r--;
while (l<r && OnRight(q[r],GetIntersection(q[l],q[l+1]))) l++;
if (r-l<=1 ) return 0;
q[r+1]=q[l];
m=0;
for (int i=l;i<=r;i++) poly[m++]=GetIntersection(q[i],q[i+1]);
return m;
}

凸包

int ConvexHull(Point* p,int n,Point* ch)
{
sort(p,p+n);
int m=0;
for(int i=0;i<n;++i)
{
while(m>1&&((ch[m-1]-ch[m-2])^(p[i]-ch[m-2]))<=0) m--;
ch[m++]=p[i];
}
int k=m;
for(int i=n-2;i>=0;i--)
{
while(m>k&&((ch[m-1]-ch[m-2])^(p[i]-ch[m-2]))<=0) m--;
ch[m++]=p[i];
}
if(n>1) m--;
return m;
}

基础模板补充：

double Cross/*B-A和C-A的叉积*/(Point A, Point B,Point C)
{
return (B-A)^(C-A);
}

double dis_pair_seg/*两条线段间的最短距离*/(Point p1, Point p2, Point p3, Point p4)
{
return min(min(DistanceToSegment(p1, p3, p4), DistanceToSegment(p2, p3, p4)),
min(DistanceToSegment(p3, p1, p2), DistanceToSegment(p4, p1, p2)));
}

旋转卡壳

double rotating_calipers(Point *ch,int n)
{
int q=1;
double ans=0;
ch
=ch[0];
for(int p=0; p<n; p++)
{
while(((ch[q+1]-ch[p+1])^(ch[p]-ch[p+1]))>((ch[q]-ch[p+1])^(ch[p]-ch[p+1]))) q=(q+1)%n;
ans=max(ans,max(Length(ch[p]-ch[q]),Length(ch[p+1]-ch[q+1])));
}
return ans;
}

为什么是大于号而不是大于等于号？因为ch默认是凸多边形

转角法判定点P是否在多边形内部（多边形不一定是凸多边形）

int isPointInPolygon(Point P, Point* Poly, int n)//转角法判定点P是否在多边形内部
{
int wn=0;
for(int i = 0; i < n; ++i)
{
if(OnSegment(P, Poly[i], Poly[(i+1)%n]))    return -1;    //在边界上
int k = dcmp((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
}

关于圆的一些模板

圆的定义：

struct Circle{
Point c;
double r;
Circle() {}
Circle(Point c, double r) : c(c), r(r) {}
Point point(double a){
return Point(c.x + cos(a) * r, c.y + sin(a) * r);
}
void read(){
c.read();
scanf("%lf", &r);
}
};