• ACM计算几何总结


    ACM计算几何总结

    1 几何基础

    #include <cstdio>
    #include <cmath>
    using namespace std;
    const double pi = acos(-1.0);
    const double inf = 1e100;
    const double eps = 1e-6;
    int sgn(double d){
        if(fabs(d) < eps)
            return 0;
        if(d > 0)
            return 1;
        return -1;
    }
    int dcmp(double x, double y){
        if(fabs(x - y) < eps)
            return 0;
        if(x > y)
            return 1;
        return -1;
    }
    int main() {
        double x = 1.49999;
        int fx = floor(x);//向下取整函数
        int cx = ceil(x);//向上取整函数
        int rx = round(x);//四舍五入函数
        printf("%f %d %d %d
    ", x, fx, cx, rx);
        //输出结果 1.499990 1 2 1
        return  0 ;
    }

    3 点与向量

    2.1 手动实现

    点积: Dot

      a·b的几何意义为a在b上的投影长度乘以b的模长

      a·b=|a||b|cosθ,其中θ为a,b之间的夹角

      坐标表示

      a=(x1,y1) b=(x2,y2)

      a·b=x1*x2+y1*y2;

    点积的应用: Angle

    (1)判断两个向量是否垂直 a⊥b <=> a·b=0

    (2)求两个向量的夹角,点积<0为钝角,点积>0为锐角

     求模长:Length

    法向量:与单位向量垂直的向量称为单位法向量Normal

    二维叉积Cross

    两个向量a和b的叉积写作a×b(有时也被写成a∧b,避免和字母x混淆)

    两个向量的叉积是一个标量,a×b的几何意义为他们所形成的平行四边形的有向面积

    坐标表示a=(x1,y1) b=(x2,y2)

    a×b=x1y2-x2y1

    直观理解,假如b在a的左边,则有向面积为正,假如在右边则为负。假如b,a共线,则叉积为0,。

    所以叉积可以用来判断平行。

      

    向量的旋转Rotate

    a=(x,y)可以看成是x*(1,0)+y1*(0,1)

    分别旋转两个单位向量,则变成x*(cosθ,sinθ)+y1*(-sinθ,cosθ)

    struct Point{
        double x, y;
        Point(double x = 0, double y = 0):x(x),y(y){}
    };
    typedef Point Vector;
    Vector operator + (Vector A, Vector 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){
        if(a.x == b.x)
            return a.y < b.y;
        return a.x < b.x;
    }
    const double eps = 1e-6;
    int sgn(double x){
        if(fabs(x) < eps)
            return 0;
        if(x < 0)
            return -1;
        return 1;
    }
    bool operator == (const Point& a, const Point& b){
        if(sgn(a.x-b.x) == 0 && sgn(a.y-b.y) == 0)
            return true;
        return false;
    }
    double Dot(Vector A, Vector B){
        return A.x*B.x + A.y*B.y;
    }
    double Length(Vector A){
        return sqrt(Dot(A, A));
    }
    double Angle(Vector A, Vector B){
        return acos(Dot(A, B)/Length(A)/Length(B));
    }
    double Cross(Vector A, Vector B){
        return A.x*B.y-A.y*B.x;
    }
    double Area2(Point A, Point B, Point C){
        return Cross(B-A, C-A);
    }
    Vector Rotate(Vector A, double rad){//rad为弧度 且为逆时针旋转的角
        return Vector(A.x*cos(rad)-A.y*sin(rad), A.x*sin(rad)+A.y*cos(rad));
    }
    Vector Normal(Vector A){//向量A左转90°的单位法向量
        double L = Length(A);
        return Vector(-A.y/L, A.x/L);
    }
    bool ToLeftTest(Point a, Point b, Point c){
        return Cross(b - a, c - b) > 0;
    }

    2.2 复数黑科技

    #include <complex>
    using namespace std;
    typedef complex<double> Point;
    typedef Point Vector;//复数定义向量后,自动拥有构造函数、加减法和数量积
    const double eps = 1e-9;
    int sgn(double x){
        if(fabs(x) < eps)
            return 0;
        if(x < 0)
            return -1;
        return 1;
    }
    double Length(Vector A){
        return abs(A);
    }
    double Dot(Vector A, Vector B){//conj(a+bi)返回共轭复数a-bi
        return real(conj(A)*B);
    }
    double Cross(Vector A, Vector B){
        return imag(conj(A)*B);
    }
    Vector Rotate(Vector A, double rad){
        return A*exp(Point(0, rad));//exp(p)返回以e为底复数的指数
    }

    3 点与线

    3.1 直线定义

    struct Line{//直线定义
        Point v, p;
        Line(Point v, Point p):v(v), p(p) {}
        Point point(double t){//返回点P = v + (p - v)*t
            return v + (p - v)*t;
        }
    };

    3.2 求两直线交点

    //调用前需保证 Cross(v, w) != 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;
    }

    3.3 求点到直线距离

    利用叉积求面积,然后除以平行四边形的底边长,得到平行四边形的高即点到直线的距离

    //点P到直线AB距离
    double DistanceToLine(Point P, Point A, Point B){
        Vector v1 = B-A, v2 = P-A;
        return fabs(Cross(v1, v2)/Length(v1));
    }//不去绝对值,得到的是有向距离

    3.4 求点到线段距离

    比点到直线的距离稍微复杂。因为是线段,所以如果平行四边形的高在区域之外的话就不合理,这时候需要计算点到距离较近的端点的距离。

    矢量算法过程清晰,如果具有一定的空间几何基础,则是解决此类问题时应优先考虑的方法。当需要计算的数据量很大时,这种方式优势明显。

    由于矢量具有方向性,故一些方向的判断直接根据其正负号就可以得知,使得其中的一些问题得以很简单的解决。用此方法考虑,

    我们只需要找到向量 方向上的投影,具体如下:

    上面的 方向上的单位向量,其意义是给所求向量确定方向。是的两个向量的内积,

       ,其中θ为向量AP与AB之间的夹角。是向量长度。

    那么即为上图中线段AC的长度值,不带有方向性。此数值与上述表征方向的 整体构成有大小、

    有方向的新向量,即为 方向上的投影向量,C为投影点。根据得到的,由向量的方向性可知:

    故根据r值的不同,最短距离

    //点P到线段AB距离公式
    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(v2);
        if(dcmp(Dot(v1, v3)) > 0)
            return Length(v3);
        return DistanceToLine(P, A, B);
    }

    3.5 求点在直线上的投影点

    //点P在直线AB上的投影点
    Point GetLineProjection(Point P, Point A, Point B){
        Vector v = B-A;
        return A+v*(Dot(v, P-A)/Dot(v, v));
    }

    3.6 判断点是否在线段上

    //判断p点是否在线段a1a2上
    bool OnSegment(Point p, Point a1, Point a2){
        return dcmp(Cross(a1-p, a2-p)) == 0 && dcmp(Dot(a1-p, a2-p)) < 0;
    }

    3.7 判断两线段是否相交

    //判断两线段是否相交
    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);
        //if判断控制是否允许线段在端点处相交,根据需要添加
        if(!sgn(c1) || !sgn(c2) || !sgn(c3) || !sgn(c4)){
            bool f1 = OnSegment(b1, a1, a2);
            bool f2 = OnSegment(b2, a1, a2);
            bool f3 = OnSegment(a1, b1, b2);
            bool f4 = OnSegment(a2, b1, b2);
            bool f = (f1|f2|f3|f4);
            return f;
        }
        return (sgn(c1)*sgn(c2) < 0 && sgn(c3)*sgn(c4) < 0);
    }

    4 三角形

     4.1求三角形外心

    struct Point {
        double x, y;
    };
    struct Line{
        Point a, b;
    };
    Point Intersection(Line u, Line v){
        Point ret = u.a;
        double t1 = (u.a.x - v.a.x)*(v.a.y - v.b.y) - (u.a.y - v.a.y)*(v.a.x - v.b.x);
        double t2 =    (u.a.x - u.b.x)*(v.a.y - v.b.y) - (u.a.y - u.b.y)*(v.a.x - v.b.x);
        double t = t1/t2;
        ret.x += (u.b.x - u.a.x)*t;
        ret.y += (u.b.y - u.a.y)*t;
        return ret;
    }
    //外心
    Point Circumcenter(Point a, Point b, Point c){
        Line u, v;
        u.a.x = (a.x + b.x)/2;
        u.a.y = (a.y + b.y)/2;
        u.b.x = u.a.x - a.y + b.y;
        u.b.y = u.a.y + a.x - b.x;
        v.a.x = (a.x + c.x)/2;
        v.a.y = (a.y + c.y)/2;
        v.b.x = v.a.x - a.y + c.y;
        v.b.y = v.a.y + a.x - c.x;
        return Intersection(u, v);
    }

    4.1 求三角形内心

    struct Point {
        double x, y;
    };
    struct Line{
        Point a, b;
    };
    Point Intersection(Line u, Line v){
        Point ret = u.a;
        double t1 = (u.a.x - v.a.x)*(v.a.y - v.b.y) - (u.a.y - v.a.y)*(v.a.x - v.b.x);
        double t2 =    (u.a.x - u.b.x)*(v.a.y - v.b.y) - (u.a.y - u.b.y)*(v.a.x - v.b.x);
        double t = t1/t2;
        ret.x += (u.b.x - u.a.x)*t;
        ret.y += (u.b.y - u.a.y)*t;
        return ret;
    }
    //三角形内心
    Point Incenter(Point a, Point b, Point c){
        Line u, v;
        double m, n;
        u.a = a;
        m = atan2(b.y - a.y, b.x - a.x);
        n = atan2(c.y - a.y, c.x - a.x);
        u.b.x = u.a.x + cos((m + n)/2);
        u.b.y = u.a.y + sin((m + n)/2);
        v.a = b;
        m = atan2(a.y - b.y, a.x - b.x);
        n = atan2(c.y - b.y, c.x - b.x);
        v.b.x = v.a.x + cos((m + n)/2);
        v.b.y = v.a.y + sin((m + n)/2);
        return Intersection(u, v);
    }

    4.2 求三角形垂心

    struct Point {
        double x, y;
    };
    struct Line{
        Point a, b;
    };
    Point Intersection(Line u, Line v){
        Point ret = u.a;
        double t1 = (u.a.x - v.a.x)*(v.a.y - v.b.y) - (u.a.y - v.a.y)*(v.a.x - v.b.x);
        double t2 =    (u.a.x - u.b.x)*(v.a.y - v.b.y) - (u.a.y - u.b.y)*(v.a.x - v.b.x);
        double t = t1/t2;
        ret.x += (u.b.x - u.a.x)*t;
        ret.y += (u.b.y - u.a.y)*t;
        return ret;
    }
    //三角形垂心
    Point Perpencenter(Point a, Point b, Point c){
        Line u, v;
        u.a = c;
        u.b.x = u.a.x - a.y + b.y;
        u.b.y = u.a.y + a.x - b.x;
        v.a = b;
        v.b.x = v.a.x - a.y + c.y;
        v.b.y = v.a.y + a.x - c.x;
        return Intersection(u, v);
    }

    4.3 求三角形重心

    struct Point {
        double x, y;
    };
    struct Line{
        Point a, b;
    };
    Point Intersection(Line u, Line v){
        Point ret = u.a;
        double t1 = (u.a.x - v.a.x)*(v.a.y - v.b.y) - (u.a.y - v.a.y)*(v.a.x - v.b.x);
        double t2 =    (u.a.x - u.b.x)*(v.a.y - v.b.y) - (u.a.y - u.b.y)*(v.a.x - v.b.x);
        double t = t1/t2;
        ret.x += (u.b.x - u.a.x)*t;
        ret.y += (u.b.y - u.a.y)*t;
        return ret;
    }
    //三角形重心
    //到三角形三顶点距离的平方和最小的点
    //三角形内到三边距离之积最大的点
    Point barycenter(Point a, Point b, Point c){
        Line u, v;
        u.a.x = (a.x + b.x)/2;
        u.a.y = (a.y + b.y)/2;
        u.b = c;
        v.a.x = (a.x + c.x)/2;
        v.a.y = (a.y + c.y)/2;
        v.b = b;
        return Intersection(u, v);
    }

    4.5 求三角形费马点

    struct Point {
        double x, y;
    };
    struct Line{
        Point a, b;
    };
    inline double Dist(Point p1, Point p2){
        return sqrt((p1.x - p2.x)*(p1.x - p2.x) + (p1.y - p2.y)*(p1.y - p2.y));
    }
    //三角形费马点
    //到三角形三顶点距离之和最小的点
    Point Ferment(Point a, Point b, Point c){
        Point u, v;
        double step = fabs(a.x) + fabs(a.y) + fabs(b.x) + fabs(b.y) + fabs(c.x) + fabs(c.y);
        u.x = (a.x + b.x + c.x)/3;
        u.y = (a.y + b.y + c.y)/3;
        while(step >1e-10){
            for(int k = 0; k < 10; step /= 2, k++){
                for(int i = -1; i <= 1; ++i){
                    for(int j = -1; j <= 1; ++j){
                        v.x = u.x + step*i;
                        v.y = u.y + step*j;
                        double t1 = Dist(u, a) + Dist(u, b) + Dist(u, c);
                        double t2 = Dist(v, a) + Dist(v, b) + Dist(v, c);
                        if (t1 > t2) u = v;
                    }
                }
            }
        }
        return u;
    }

    5 多边形

    5.1 求多边形有向面积

    //多边形有向面积
    double PolygonArea(Point* p, int n){//p为端点集合,n为端点个数
        double s = 0;
        for(int i = 1; i < n-1; ++i)
            s += Cross(p[i]-p[0], p[i+1]-p[0]);
        return s;
    }

    5.2 判断点是否在多边形内

    //判断点是否在多边形内,若点在多边形内返回1,在多边形外部返回0,在多边形上返回-1
    int isPointInPolygon(Point p, vector<Point> poly){
        int wn = 0;
        int n = poly.size();
        for(int i = 0; i < n; ++i){
            if(OnSegment(p, poly[i], poly[(i+1)%n])) return -1;
            int k = sgn(Cross(poly[(i+1)%n] - poly[i], p - poly[i]));
            int d1 = sgn(poly[i].y - p.y);
            int d2 = sgn(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;
        return 0;
    }

    6 圆

    6.1 求圆与直线交点

    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;//判别式
        if(sgn(delta) < 0)//相离
            return 0;
        if(sgn(delta) == 0){//相切
            t1 = -f /(2*e);
            t2 = -f /(2*e);
            sol.push_back(L.point(t1));//sol存放交点本身
            return 1;
        }
        //相交
        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;
    }

    6.2 求两圆交点

    6.3 求点到圆的切线

    6.4 求两圆公切线

    6.5 求两圆相交面积

    double AreaOfOverlap(Point c1, double r1, Point c2, double r2){
        double d = Length(c1 - c2);
        if(r1 + r2 < d + eps)
            return 0.0;
        if(d < fabs(r1 - r2) + eps){
            double r = min(r1, r2);
            return pi*r*r;
        }
        double x = (d*d + r1*r1 - r2*r2)/(2.0*d);
        double p = (r1 + r2 + d)/2.0;
        double t1 = acos(x/r1);
        double t2 = acos((d - x)/r2);
        double s1 = r1*r1*t1;
        double s2 = r2*r2*t2;
        double s3 = 2*sqrt(p*(p - r1)*(p - r2)*(p - d));
        return s1 + s2 - s3;
    }

    6.6 用给定半径的圆覆盖最多的点

    const int maxn = 3e2 + 5;
    const double PI = acos(-1.0);
    struct Point{
        double x, y;
    }p[maxn];
    struct Angle{
        double pos;
        bool in;
        bool operator < (const Angle& a) const {
            return pos < a.pos;
        }
    }a[maxn<<1];
    inline double Dist(Point& p1, Point& p2){
        return sqrt((p1.x - p2.x)*(p1.x - p2.x) + (p1.y - p2.y)*(p1.y - p2.y));
    }
    //点集的大小n,给定圆的半径r,返回最多覆盖的点的个数
    int solve(int n, double r){
        int ret = 1;
        for(int i = 0; i < n; ++i){
            int m = 0;
            for(int j = 0; j < n; ++j){
                if(i == j) continue;
                double d = Dist(p[i], p[j]);
                if(d > 2*r) continue;
                double alpha = atan2(p[j].y - p[i].y, p[j].x - p[i].x);
                double beta = acos(d/(2*r));
                a[m].pos = alpha - beta;
                a[m++].in = true;
                a[m].pos = alpha + beta;
                a[m++].in = false;
            }
            sort(a, a + m);
            int t = 1;
            for(int j = 0; j < m; ++j){
                if(a[j].in) ++t;
                else --t;
                if(ret < t) ret = t;
            }
        }
        return ret;
    }

    7 凸包

    7.1 GrahamScan O(nlogn)

    const int maxn = 1e3 + 5;
    const double eps = 1e-9;
    struct Point {
        double x, y;
        Point(double x = 0, double y = 0):x(x),y(y){}
    };
    typedef Point Vector;
    Point lst[maxn];
    int stk[maxn], top;
    Vector operator - (Point A, Point B){
        return Vector(A.x-B.x, A.y-B.y);
    }
    int sgn(double x){
        if(fabs(x) < eps)
            return 0;
        if(x < 0)
            return -1;
        return 1;
    }
    double Cross(Vector v0, Vector v1) {
        return v0.x*v1.y - v1.x*v0.y;
    }
    double Dis(Point p1, Point p2) { //计算 p1p2的 距离
        return sqrt((p2.x-p1.x)*(p2.x-p1.x)+(p2.y-p1.y)*(p2.y-p1.y));
    }
    bool cmp(Point p1, Point p2) { //极角排序函数 ,角度相同则距离小的在前面
        int tmp = sgn(Cross(p1 - lst[0], p2 - lst[0]));
        if(tmp > 0)
            return true;
        if(tmp == 0 && Dis(lst[0], p1) < Dis(lst[0], p2))
            return true;
        return false;
    }
    //点的编号0 ~ n - 1
    //返回凸包结果stk[0 ~ top - 1]为凸包的编号
    void Graham(int n) {
        int k = 0;
        Point p0;
        p0.x = lst[0].x;
        p0.y = lst[0].y;
        for(int i = 1; i < n; ++i) {
            if( (p0.y > lst[i].y) || ((p0.y == lst[i].y) && (p0.x > lst[i].x)) ) {
                p0.x = lst[i].x;
                p0.y = lst[i].y;
                k = i;
            }
        }
        lst[k] = lst[0];
        lst[0] = p0;
        sort(lst + 1, lst + n, cmp);
        if(n == 1) {
            top = 1;
            stk[0] = 0;
            return ;
        }
        if(n == 2) {
            top = 2;
            stk[0] = 0;
            stk[1] = 1;
            return ;
        }
        stk[0] = 0;
        stk[1] = 1;
        top = 2;
        for(int i = 2; i < n; ++i) {
            while(top > 1 && Cross(lst[stk[top - 1]] - lst[stk[top - 2]], lst[i] - lst[stk[top - 2]]) <= 0)
                --top;
            stk[top] = i;
            ++top;
        }
        return ;
    }

    7.2 Andrew O(nlogn)

    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);
    }
    bool operator < (const Point& a, const Point& b){
        if(a.x == b.x)
            return a.y < b.y;
        return a.x < b.x;
    }
    double Cross(Vector v0, Vector v1) {
        return v0.x*v1.y - v1.x*v0.y;
    }
    //计算凸包,输入点数组为 p,个数为 n, 输出点数组为 ch。函数返回凸包顶点数
    //如果不希望凸包的边上有输入点,则把两个 <= 改为 <
    //在精度要求高时建议用dcmp比较
    //输入不能有重复点,函数执行完后输入点的顺序被破坏
    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 && Cross(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 && Cross(ch[m-1] - ch[m-2], p[i] - ch[m-2]) <= 0) m--;
            ch[m++] = p[i];
        }
        if(n > 1) --m;
        return m;
    }

    8 半平面交 O(nlogn)

    const double eps = 1e-6;
    struct Point{
        double x, y;
        Point(double x = 0, double y = 0):x(x),y(y){}
    };
    typedef Point Vector;
    Vector operator + (Vector A, Vector 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);
    }
    int sgn(double x){
        if(fabs(x) < eps)
            return 0;
        if(x < 0)
            return -1;
        return 1;
    }
    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 Length(Vector A){
        return sqrt(Dot(A, A));
    }
    Vector Normal(Vector A){//向量A左转90°的单位法向量
        double L = Length(A);
        return Vector(-A.y/L, A.x/L);
    }
    struct Line{
        Point p;//直线上任意一点
        Vector v;//方向向量,它的左边就是对应的半平面
        double ang;//极角,即从x轴正半轴旋转到向量v所需要的角(弧度)
        Line(){}
        Line(Point p, Vector v) : p(p), v(v){
            ang = atan2(v.y, v.x);
        }
        bool operator < (const Line& L) const {//排序用的比较运算符
            return ang < L.ang;
        }
    };
    //点p在有向直线L的左侧
    bool OnLeft(Line L, Point p){
        return Cross(L.v, p - L.p) > 0;
    }
    //两直线交点。假定交点唯一存在
    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;
    }
    //半平面交的主过程
    int HalfplaneIntersection(Line* L, int n, Point* poly){
        sort(L, L + n);//按照极角排序
        int fst = 0, lst = 0;//双端队列的第一个元素和最后一个元素
        Point *P = new Point[n];//p[i] 为 q[i]与q[i + 1]的交点
        Line *q = new Line[n];//双端队列
        q[fst = lst = 0] = L[0];//初始化为只有一个半平面L[0]
        for(int i = 1; i < n; ++i){
            while(fst < lst && !OnLeft(L[i], P[lst - 1])) --lst;
            while(fst < lst && !OnLeft(L[i], P[fst])) ++fst;
            q[++lst] = L[i];
            if(sgn(Cross(q[lst].v, q[lst - 1].v)) == 0){
                //两向量平行且同向,取内侧一个
                --lst;
                if(OnLeft(q[lst], L[i].p)) q[lst] = L[i];
            }
            if(fst < lst)
                P[lst - 1] = GetIntersection(q[lst - 1], q[lst]);
        }
        while(fst < lst && !OnLeft(q[fst], P[lst - 1])) --lst;
        //删除无用平面
        if(lst - fst <= 1) return 0;//空集
        P[lst] = GetIntersection(q[lst], q[fst]);//计算首尾两个半平面的交点
        //从deque复制到输出中
        int m = 0;
        for(int i = fst; i <= lst; ++i) poly[m++] = P[i];
        return m;
    }

    9 平面最近点对 O(nlogn)

    struct Point {
        double x,y;
        bool operator <(const Point &a)const {
            return x < a.x;
        }
    };
    inline double dist(const Point &p1, const Point &p2) {
        return sqrt((p1.x - p2.x) * (p1.x - p2.x) + (p1.y - p2.y) * (p1.y - p2.y));
    }
    Point p[maxn], q[maxn];
    double ClosestPair(int l, int r) {
        if(l == r)
            return inf;
        int mid = (l+r)>>1;
        double tx = p[mid].x;
        int tot = 0;
        double ret = min(ClosestPair(l, mid), ClosestPair(mid + 1, r));
        for(int i = l, j = mid + 1; (i <= mid || j <= r); ++i) {
            while(j <= r && (p[i].y > p[j].y || i > mid)) {
                q[tot++] = p[j];
                j++; //归并按y排序
            }
            if(abs(p[i].x - tx) < ret && i <= mid) { //选择中间符合要求的点
                for(int k = j - 1; k > mid && j - k < 3; --k)
                    ret = min(ret, dist(p[i], p[k]));
                for(int k = j; k <= r && k-j < 2; ++k)
                    ret = min(ret, dist(p[i], p[k]));
            }
            if(i <= mid)
                q[tot++] = p[i];
        }
        for(int i = l, j = 0; i <= r; ++i, ++j)
            p[i] = q[j];
        return ret;
    }

    10 旋转卡壳

    10.1 求凸包直径

    double Dist2(Point p1, Point p2) { //计算距离的平方
        double ret = Dot(p1 - p2, p1 - p2);
        return ret;
    }
    double RotatingCalipers(Point* ch, int m) {//返回平面最大距离的平方
        if(m == 1) return 0.0;
        if(m == 2) return Dist2(ch[0], ch[1]);
        double ret = 0.0;
        ch[m] = ch[0];
        int j = 2;
        for(int i = 0; i < m; ++i) {
            while(Cross(ch[i + 1] - ch[i], ch[j] - ch[i]) < Cross(ch[i + 1] - ch[i], ch[j + 1] - ch[i]))
                j = (j + 1)%m;
            ret = max(ret, max(Dist2(ch[j], ch[i]), Dist2(ch[j], ch[i + 1])));
        }
        return ret;
    }

    10.2 求凸包的宽(凸包对踵点的最小值)

    double dist(Point a, Point b){
        double ret = sqrt((a.x - b.x)*(a.x - b.x) + (a.y - b.y)*(a.y - b.y));
        return ret;
    }
    double mxid[maxn];
    double RotateCalipersWide(Point* p, int n){
        int j = 1;
        p[n] = p[0];
        for(int i = 0; i < n; ++i){
            while(fabs(Cross(p[i + 1] - p[i], p[j + 1] - p[i])) > fabs(Cross(p[i + 1] - p[i], p[j] - p[i]))) j = (j + 1)%n;
            mxid[i] = fabs(Cross(p[j] - p[i], p[i + 1] - p[i]))/dist(p[i], p[i + 1]);
        }
        double ret = 1e18;
        for(int i = 0; i < n; ++i) ret = min(ret, mxid[i]);
        return ret;
    }

    11 网格图

    11.1 Pick定理

    11.2 多边形与网格点

    struct Point{
        int x, y;
    };
    //多边形上的网格点个数
    int Onedge(int n, Point* p){
        int ret = 0;
        for(int i = 0; i < n; ++i)
            ret += __gcd(abs(p[i].x - p[(i + 1)%n].x), abs(p[i].y - p[(i + 1)%n].y));
        return ret;
    }
    //多边形内的网格点个数
    int Inside(int n, Point* p){
        int ret = 0;
        for (int i = 0; i < n; ++i)
            ret += p[(i + 1)%n].y*(p[i].x - p[(i + 2)%n].x);
        ret = (abs(ret) - Onedge(n, p))/2 + 1;
        return ret;
    }

    相关博客:

    https://blog.csdn.net/linxilinxilinxi/article/details/82624098?utm_medium=distribute.pc_relevant_t0.none-task-blog-BlogCommendFromMachineLearnPai2-1.add_param_isCf&depth_1-utm_source=distribute.pc_relevant_t0.none-task-blog-BlogCommendFromMachineLearnPai2-1.add_param_isCf

    https://blog.csdn.net/HW140701/article/details/53310003?utm_medium=distribute.pc_relevant_download.none-task-blog-baidujs-1.nonecase&depth_1-utm_source=distribute.pc_relevant_download.none-task-blog-baidujs-1.nonecase

    https://blog.csdn.net/clover_hxy/article/details/53966405

    因上求缘,果上努力~~~~ 作者:每天卷学习,转载请注明原文链接:https://www.cnblogs.com/BlairGrowing/p/13928863.html

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  • 原文地址:https://www.cnblogs.com/BlairGrowing/p/13928863.html
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