lrj计算几何模板

整理了一下大白书上的计算几何模板。

 #include <cstdio>
 #include <algorithm>
 #include <cmath>
 #include <vector>
 using namespace std;
 //lrj计算几何模板
 struct Point
 {
     double x, y;
     Point(double x=, double y=) :x(x),y(y) {}
 };
 typedef Point Vector;

 Point read_point(void)
 {
     double x, y;
     scanf("%lf%lf", &x, &y);
     return Point(x, y);
 }

 const double EPS = 1e-;

 //向量+向量=向量 点+向量=点
 Vector operator + (Vector A, Vector B)    { return Vector(A.x + B.x, A.y + B.y); }

 //向量-向量=向量 点-点=向量
 Vector operator - (Vector A, Vector B)    { return Vector(A.x - B.x, A.y - B.y); }

 //向量*数=向量
 Vector operator * (Vector A, double p)    { return Vector(A.x*p, A.y*p); }

 //向量/数=向量
 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); }

 int dcmp(double x)
 { if(fabs(x) < EPS) return ; else return x <  ? - : ; }

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

 /**********************基本运算**********************/

 //点积
 double Dot(Vector A, Vector B)
 { return A.x*B.x + A.y*B.y; }
 //向量的模
 double Length(Vector A)    { return sqrt(Dot(A, A)); }

 //向量的夹角，返回值为弧度
 double Angle(Vector A, Vector B)
 { return acos(Dot(A, B) / Length(A) / Length(B)); }

 //叉积
 double Cross(Vector A, Vector B)
 { return A.x*B.y - A.y*B.x; }

 //向量AB叉乘AC的有向面积
 double Area2(Point A, Point B, Point C)
 { return Cross(B-A, C-A); }

 //向量A旋转rad弧度
 Vector VRotate(Vector A, double rad)
 {
     return Vector(A.x*cos(rad) - A.y*sin(rad), A.x*sin(rad) + A.y*cos(rad));
 }

 //将B点绕A点旋转rad弧度
 Point PRotate(Point A, Point B, double rad)
 {
     return A + VRotate(B-A, rad);
 }

 //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
 Vector Normal(Vector A)
 {
     double l = Length(A);
     return Vector(-A.y/l, A.x/l);
 }

 /**********************点和直线**********************/

 //求直线P + tv 和 Q + tw的交点，调用前要确保两条直线有唯一交点
 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;
 }//在精度要求极高的情况下，可以自定义分数类

 //P点到直线AB的距离
 double DistanceToLine(Point P, Point A, Point B)
 {
     Vector v1 = B - A, v2 = P - A;
     return fabs(Cross(v1, v2)) / Length(v1);    //不加绝对值是有向距离
 }

 //点到线段的距离
 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)) < )    return Length(v2);
     else if(dcmp(Dot(v1, v3)) > )    return Length(v3);
     else return fabs(Cross(v1, v2)) / Length(v1);
 }

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

 //线段“规范”相交判定
 bool SegmentProperIntersection(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)< && dcmp(c3)*dcmp(c4)<;
 }

 //判断点是否在线段上
 bool OnSegment(Point P, Point a1, Point a2)
 {
     Vector v1 = a1 - P, v2 = a2 - P;
     return dcmp(Cross(v1, v2)) ==  && dcmp(Dot(v1, v2)) < ;
 }

 //求多边形面积
 double PolygonArea(Point* P, int n)
 {
     double ans = 0.0;
     for(int i = ; i < n - ; ++i)
         ans += Cross(P[i]-P[], P[i+]-P[]);
     return ans/;
 }

 int main(void)
 {
     Vector a[];
     sort(a, a + );
     return ;
 }

 /**********************圆的相关计算**********************/

 const double PI = acos(-1.0);
 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
     {
         return ang < L.ang;
     }
 };

 struct Circle
 {
     Point c;    //圆心
     double r;    //半径
     Circle(Point c, double r):c(c), r(r)    {}
     Point point(double a)
     {//求对应圆心角的点
         return Point(c.x + r*cos(a), c.y + r*sin(a));
     }
 };

 //两圆相交并返回交点个数
 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 = *(a*b + c*d), g = b*b + d*d - C.r*C.r;
     double delta = f*f - *e*g;        //判别式
     if(dcmp(delta) < )    return ;    //相离
     if(dcmp(delta) == )            //相切
     {
         t1 = t2 = -f / ( * e);
         sol.push_back(L.point(t1));
         return ;
     }
     //相交
     t1 = (-f - sqrt(delta)) / ( * e);    sol.push_back(L.point(t1));
     t2 = (-f + sqrt(delta)) / ( * e);    sol.push_back(L.point(t2));
     return ;
 }

 //计算向量极角
 double angle(Vector v)    { return atan2(v.y, v.x); }

 int getCircleCircleIntersection(Circle C1, Circle C2, vector<Point>& sol)
 {//圆与圆相交，并返回交点个数
     double d = Length(C1.c - C2.c);
     if(dcmp(d) == )
     {
         if(dcmp(C1.r - C2.r) == )    return -;    //两圆重合
         return ;                                //没有交点
     }
     if(dcmp(C1.r + C2.r - d) > )    return ;
     if(dcmp(fabs(C1.r - C2.r) - d) > )    return ;

     double a = angle(C2.c - C1.c);
     double da = acos((C1.r*C1.r + d*d - C2.r*C2.r) / (*C1.r*d));
     Point p1 = C1.point(a+da), p2 = C1.point(a-da);
     sol.push_back(p1);
     if(p1 == p2)    return ;
     sol.push_back(p2);
     return ;
 }

 //过定点作圆的切线并返回切线条数
 int getTangents(Point p, Circle C, Vector* v)
 {
     Vector u = C.c - p;
     double dist = Length(u);
     if(dist < C.r)    return ;
     else if(dcmp(dist - C.r) == )
     {
         v[] = VRotate(u, PI/);
         return ;
     }
     else
     {
         double ang = asin(C.r / dist);
         v[] = VRotate(u, +ang);
         v[] = VRotate(u, -ang);
         return ;
     }
 }

 //求两个圆的公切线，并返回切线条数
 //注意，这里的Circle和上面的定义的Circle不一样
 int getTangents(Circle A, Circle B, Point* a, Point* b)
 {
     int cnt = ;
     if(A.r < B.r)    { swap(A, B); swap(a, b); }
     double d2 = (A.x-B.x)*(A.x-B.x) + (A.y-B.y)*(A.y-B.y);
     double rdiff = A.r - B.r;
     double rsum = A.r + B.r;
     if(d2 < rdiff*rdiff)    return ;    //内含

     double base = atan2(B.y-A.y, B.x-A.x);
     if(dcmp(d2) ==  && dcmp(A.r - B.r) == )    return -; //重合
     if(dcmp(d2 - rdiff*rdiff) == )    //内切
     {
         a[cnt] = A.point(base); b[cnt] = B.point(base); cnt++;
         return ;
     }

     //有外公切线
     double ang = acos((A.r - B.r) / sqrt(d2));
     a[cnt] = A.point(base + ang); b[cnt] = B.point(base + ang); cnt++;
     a[cnt] = A.point(base - ang); b[cnt] = B.point(base - ang); cnt++;
     if(dcmp(rsum*rsum - d2) == )
     {//外切
         a[cnt] = b[cnt] = A.point(base); cnt++;
     }
     else if(dcmp(d2 - rsum*rsum) > )
     {
         ang = acos((A.r + B.r) / sqrt(d2));
         a[cnt] = A.point(base + ang); b[cnt] = B.point(PI + base + ang); cnt++;
         a[cnt] = A.point(base - ang); b[cnt] = B.point(PI + base - ang); cnt++;
     }
     return cnt;
 }

 //转角发判定点P是否在多边形内部
 int isPointInPolygon(Point P, Point* Poly, int n)
 {
     int wn;
     for(int i = ; i < n; ++i)
     {
         if(OnSegment(P, Poly[i], Poly[(i+)%n]))    return -;    //在边界上
         int k = dcmp(Cross(Poly[(i+)%n] - Poly[i], P - Poly[i]));
         int d1 = dcmp(Poly[i].y - P.y);
         int d2 = dcmp(Poly[(i+)%n].y - P.y);
         if(k >  && d1 <=  && d2 > )    wn++;
         if(k <  && d2 <=  && d1 > )    wn--;
     }
     if(wn != )    return ;    //内部
     return ;                //外部
 }

 //计算凸包，输入点数组P，个数为n，输出点数组ch。函数返回凸包顶点数。
 //输入不能有重复点，函数执行后点的顺序会发生变化
 //如果不希望凸包的边上有输入点，把两个 <= 改成 <
 //在精度要求高时，可用dcmp比较
 int ConvexHull(Point* p, int n, Point* ch)
 {
     sort(p, p +n);
     int m = ;
     for(int i = ; i < n; ++i)
     {
         while(m >  && Cross(ch[m-] - ch[m-], p[i] - ch[m-]) <= )    m--;
         ch[m++] = p[i];
     }
     int k = m;
     for(int i = n-; i >= ; --i)
     {
         while(m > k && Cross(ch[m-] - ch[m-], p[i] - ch[m-]) <= )    m--;
         ch[m++] = p[i];
     }
     if(n > )    m--;
     return m;
 }

旋转卡壳的模板：

int diameter2(vector<Point>& points)
{
    vector<Point> p = ConvexHull(points);
    int n = p.size();
    //for(int i = 0; i < n; ++i)    printf("%d %d\n", p[i].x, p[i].y);
    if(n == )    return ;
    if(n == )  return Dist2(p[], p[]);
    p.push_back(p[]);
    int ans = ;
    for(int u = , v = ; u < n; ++u)
    {// 一条直线贴住边p[u]-p[u+1]
        while(true)
        {
            // 当Area(p[u], p[u+1], p[v+1]) <= Area(p[u], p[u+1], p[v])时停止旋转
            //因为两个三角形有一公共边，所以面积大的那个点到直线距离大
            // 即Cross(p[u+1]-p[u], p[v+1]-p[u]) - Cross(p[u+1]-p[u], p[v]-p[u]) <= 0
            // 根据Cross(A,B) - Cross(A,C) = Cross(A,B-C)
            // 化简得Cross(p[u+1]-p[u], p[v+1]-p[v]) <= 0
            int diff = Cross(p[u+]-p[u], p[v+]-p[v]);
            if(diff <= )
            {
                ans = max(ans, Dist2(p[u], p[v]));
                if(diff == )    ans = max(ans, Dist2(p[u], p[v+]));
                break;
            }
            v = (v+)%n;
        }
    }
    return ans;
}