【模板】计算几何

1 二维向量/点、计算几何基础

const double eps = 1e-8;

#define lt(x, y) ((x) < (y) - eps)
#define gt(x, y) ((x) > (y) + eps)
#define le(x, y) ((x) <= (y) + eps)
#define ge(x, y) ((x) >= (y) - eps)
#define eq(x, y) (le(x, y) && ge(x, y))
#define dot(x, y, z) (((y) - (x)) * ((z) - (x)))
#define cross(x, y, z) (((y) - (x)) ^ ((z) - (x)))

struct vec2 {
    double x, y;
    vec2 (double x0 = 0.0, double y0 = 0.0) : x(x0), y(y0) {}
    vec2 * read() {scanf("%lf%lf", &x, &y); return this;}
    inline vec2 operator - () const {return vec2(-x, -y);}
    inline vec2 operator + (const vec2 &B) const {return vec2(x + B.x, y + B.y);}
    inline vec2 operator - (const vec2 &B) const {return vec2(x - B.x, y - B.y);}
    inline vec2 operator * (double k) const {return vec2(x * k, y * k);}
    inline vec2 operator / (double k) const {return *this * (1.0 / k);}
    inline double operator * (const vec2 &B) const {return x * B.x + y * B.y;}
    inline double operator ^ (const vec2 &B) const {return x * B.y - y * B.x;}
    inline double norm2() const {return x * x + y * y;}
    inline double norm() const {return sqrt(x * x + y * y);}
    inline bool operator < (const vec2 &B) const {return lt(x, B.x) || le(x, B.x) && lt(y, B.y);}
    inline bool operator == (const vec2 &B) const {return eq(x, B.x) && eq(y, B.y);}
    inline bool operator << (const vec2 &B) const {return lt(y, 0) ^ lt(B.y, 0) ? lt(B.y, 0) : gt(*this ^ B, 0) || ge(*this ^ B, 0) && ge(x, 0) && lt(B.x, 0);}
    inline vec2 trans(double a11, double a12, double a21, double a22) const {return vec2(x * a11 + y * a12, x * a21 + y * a22);}
};

/*
operator * : Dot product
operator ^ : Cross product
norm2() : |v|^2 = v.v
norm() : |v| = sqrt(v.v)
operator < : Two-key compare
operator << : Polar angle compare
trans : Transition with a 2x2 matrix
*/

2 直线及其函数

struct line {
    double A, B, C; // Ax + By + C = 0, > 0 if it represents halfplane.
    line (double A0 = 0.0, double B0 = 0.0, double C0 = 0.0) : A(A0), B(B0), C(C0) {}
    line (const vec2 &u, const vec2 &v) : A(u.y - v.y), B(v.x - u.x), C(u ^ v) {} // left > 0
    inline vec2 normVec() const {return vec2(A, B);}
    inline double norm2() const {return A * A + B * B;}
    inline double operator () (const vec2 &P) const {return A * P.x + B * P.y + C;}
};

inline vec2 intersection(const line u, const line v) {return vec2(u.B * v.C - u.C * v.B, u.C * v.A - u.A * v.C) / (u.A * v.B - u.B * v.A);}

inline bool parallel(const line u, const line v) {double p = u.normVec() ^ v.normVec(); return eq(p, 0);}

inline bool perpendicular(const line u, const line v) {double p = u.normVec() * v.normVec(); return eq(p, 0);}

inline bool sameDir(const line u, const line v) {return parallel(u, v) && gt(u.normVec() * v.normVec(), 0);}

inline line bisector(const vec2 u, const vec2 v) {return line(v.x - u.x, v.y - u.y, 0.5 * (u.norm2() - v.norm2()));}

inline double dis2(const vec2 P, const line l) {return l(P) * l(P) / l.norm2();}

inline vec2 projection(const vec2 P, const line l) {return P - l.normVec() * (l(P) / l.norm2());}

inline vec2 symmetry(const vec2 P, const line l) {return P - l.normVec() * (2 * l(P) / l.norm2());}

3 多边形操作

// Relation of 3 points. (2 inside, 1 outside, 0 not collinear)
inline int collinear(const vec2 u, const vec2 v, const vec2 P) {double p = cross(P, u, v); return eq(p, 0) ? 1 + le(dot(P, u, v), 0) : 0;}

// Perimeter of a polygon
double perimeter(int n, vec2 *poly) {
    double ret = (poly[n - 1] - *poly).norm();
    for (int i = 1; i < n; ++i) ret += (poly[i - 1] - poly[i]).norm();
    return ret;
}

// Directed area of a polygon (positive if CCW)
double area(int n, vec2 *poly) {
    double ret = poly[n - 1] ^ *poly;
    for (int i = 1; i < n; ++i) ret += poly[i - 1] ^ poly[i];
    return ret * 0.5;
}

// Point in polygon (2 on boundary, 1 inside, 0 outside)
int contain(int n, vec2 *poly, const vec2 P) {
    int in = 0; vec2 *r = poly + (n - 1), *u, *v;
    for (int i = 0; i < n; r = poly, ++poly, ++i) {
        if (collinear(*r, *poly, P) == 2) return 2;
        gt(r->y, poly->y) ? (u = poly, v = r) : (u = r, v = poly);
        if (ge(P.y, v->y) || lt(P.y, u->y)) continue;
        in ^= gt(cross(P, *u, *v), 0);
    }
    return in;
}

4 平面凸包 (Graham Scan)

int graham(int n, vec2 *p, vec2 *dest) {
    int i; vec2 *ret = dest;
    std::iter_swap(p, std::min_element(p, p + n));
    std::sort(p + 1, p + n, [p] (const vec2 A, const vec2 B) {double r = cross(*p, A, B); return gt(r, 0) || (ge(r, 0) && lt((A - *p).norm2(), (B - *p).norm2()));});
    for (i = 0; i < 2 && i < n; ++i) *ret++ = p[i];
    for (; i < n; *ret++ = p[i++])
        for (; ret != dest + 1 && ge(cross(ret[-2], p[i], ret[-1]), 0); --ret);
    return *ret = *p, ret - dest;
}

5 旋转卡壳求凸集直径

double convDiameter(int n, vec2 *poly) {
    int l = std::min_element(poly, poly + n) - poly, r = std::max_element(poly, poly + n) - poly, i = l, j = r;
    double diam = (poly[l] - poly[r]).norm2();
    do {
        (ge(poly[(i + 1) % n] - poly[i] ^ poly[(j + 1) % n] - poly[j], 0) ? ++j : ++i) %= n;
        up(diam, (poly[i] - poly[j]).norm2());
    } while (i != l || j != r);
    return diam;
}

6 三角形外心 & 最小圆覆盖

inline vec2 circumCenter(const vec2 A, const vec2 B, const vec2 C) {
    vec2 a = B - A, b = C - A, AO;
    double det = 0.5 / (a ^ b), na = a.norm2(), nb = b.norm2();
    AO = vec2((na * b.y - nb * a.y) * det, (nb * a.x - na * b.x) * det);
    return A + AO;
}

double minCircleCover(int n, vec2 *p, vec2 *ret = NULL) {
    int i, j, k; double r2 = 0.0;
    std::random_shuffle(p + 1, p + (n + 1));
    vec2 C = p[1];
    for (i = 2; i <= n; ++i)
        if (gt((p[i] - C).norm2(), r2))
            for (C = p[i], r2 = 0, j = 1; j < i; ++j)
                if (gt((p[j] - C).norm2(), r2))
                    for (C = (p[i] + p[j]) * 0.5, r2 = (p[j] - C).norm2(), k = 1; k < j; ++k)
                        if (gt((p[k] - C).norm2(), r2))
                            C = circumCenter(p[i], p[j], p[k]), r2 = (p[k] - C).norm2();
    return ret ? *ret = C : 0, r2;
}

7 半平面交 (平行处理版)

inline bool HPIcmp(const line u, const line v) {return sameDir(u, v) ? gt((fabs(u.A) + fabs(u.B)) * v.C, (fabs(v.A) + fabs(v.B)) * u.C) : u.normVec() << v.normVec();}

inline bool geStraight(const vec2 A, const vec2 B) {return lt(A ^ B, 0) || le(A ^ B, 0) && lt(A * B, 0);}

inline bool para_nega_test(const line u, const line v) {
    return parallel(u, v) && lt(u.normVec() * v.normVec(), 0) && (fabs(u.A) + fabs(u.B)) * v.C + (fabs(v.A) + fabs(v.B)) * u.C < -7e-14;
}

int HPI(int n, line *l, vec2 *poly, vec2 *&ret) { // -1 : Unbounded, -2 : Infeasible
    int h = 0, t = -1, i, j, que[n + 5];
    std::sort(l, l + n, HPIcmp);
    n = std::unique(l, l + n, sameDir) - l;
    for (j = i = 0; i < n && j < n; ++i) {
        for (up(j, i + 1); j < n && !geStraight(l[i].normVec(), l[j].normVec()); ++j);
        if (para_nega_test(l[i], l[j])) return -2;
    }
    if (n <= 2 || geStraight(l[n - 1].normVec(), l->normVec())) return -1;
    for (i = 0; i < n; ++i) {
        if (geStraight(l[que[t]].normVec(), l[i].normVec())) return -1;
        for (; h < t && le(l[i](poly[t - 1]), 0); --t);
        for (; h < t && le(l[i](poly[h]), 0); ++h);
        que[++t] = i; h < t ? poly[t - 1] = intersection(l[ que[t - 1] ], l[ que[t] ]) : 0;
    }
    for (; h < t && le(l[ que[h] ](poly[t - 1]), 0); --t);
    return h == t ? -2 : (poly[t] = intersection(l[ que[t] ], l[ que[h] ]), ret = poly + h, t - h + 1);
}

8 平面上的圆几何

const double pi = M_PI, pi2 = pi * 2., pi_2 = M_PI_2;

inline double angle(const vec2 u, const vec2 v) {return atan2(u ^ v, u * v);}

// intersection of circle and line
int intersection(double r2, const vec2 O, const line l, vec2 *ret) {
    double d2 = dis2(O, l); vec2 j = l.normVec();
    if (gt(d2, r2)) return ret[0] = ret[1] = vec2(INFINITY, INFINITY), 0;
    if (ge(d2, r2)) return ret[0] = ret[1] = projection(O, l), 1;
    if (le(d2, 0)) {
        j = j * sqrt(r2 / j.norm2());
        ret[0] = O + j.trans(0, -1, 1, 0);
        ret[1] = O + j.trans(0, 1, -1, 0);
    } else {
        double T = copysign(sqrt((r2 - d2) / d2), l(O)); j = j * (-l(O) / j.norm2());
        ret[0] = O + j.trans(1, T, -T, 1);
        ret[1] = O + j.trans(1, -T, T, 1);
    }
    return 2;
}

// area of intersection(x^2 + y^2 = r^2, triangle OBC)
double interArea(double r2, const vec2 B, const vec2 C) {
    if (eq(B ^ C, 0)) return 0;
    vec2 is[2];
    if (!intersection(r2, vec2(), line(B, C), is)) return 0.5 * r2 * angle(B, C);
    bool b = gt(B.norm2(), r2), c = gt(C.norm2(), r2);
    if (b && c) return 0.5 * (lt(dot(*is, B, C), 0) ? r2 * (angle(B, *is) + angle(is[1], C)) + (is[0] ^ is[1]) : r2 * angle(B, C));
    else if (b) return 0.5 * (r2 * angle(B, *is) + (*is ^ C));
    else if (c) return 0.5 * ((B ^ is[1]) + r2 * angle(is[1], C));
    else return 0.5 * (B ^ C);
}

// tangents to circle((0, 0), r) through P
int tangent(double r, const vec2 P, line *ret) {
    double Q = P.norm2() - r * r;
    if (lt(Q, 0)) return ret[0] = ret[1] = line(INFINITY, INFINITY, INFINITY), 0;
    if (le(Q, 0)) return ret[0] = ret[1] = line(P.x, P.y, -P.norm2()), 1;
    Q = sqrt(Q) / r;
    ret[0] = line(P.x + Q * P.y, P.y - Q * P.x, -P.norm2());
    ret[1] = line(P.x - Q * P.y, P.y + Q * P.x, -P.norm2());
    return 2;
}

// tangets to circle(O, r) through P
int tangent(double r, const vec2 O, const vec2 P, line *ret) {
    int R = tangent(r, P - O, ret);
    if (R)
        ret[0].C -= ret[0].A * O.x + ret[0].B * O.y,
        ret[1].C -= ret[1].A * O.x + ret[1].B * O.y;
    return R;
}

9 三维计算几何基础

#define triple(x, y, z) ((x) * ((y) ^ (z)))

struct vec3 {
    double x, y, z;
    vec3 (double x0 = 0, double y0 = 0, double z0 = 0) : x(x0), y(y0), z(z0) {}
    vec3 * read() {scanf("%lf%lf%lf", &x, &y, &z); return this;}
    inline vec3 operator - () const {return vec3(-x, -y, -z);}
    inline vec3 operator + (const vec3 &B) const {return vec3(x + B.x, y + B.y, z + B.z);}
    inline vec3 operator - (const vec3 &B) const {return vec3(x - B.x, y - B.y, z - B.z);}
    inline vec3 operator * (double k) const {return vec3(x * k, y * k);}
    inline vec3 operator / (double k) const {return *this * (1.0 / k);}
    inline double operator * (const vec3 &B) const {return x * B.x + y * B.y + z * B.z;}
    inline vec3 operator ^ (const vec3 &B) const {return vec3(y * B.z - z * B.y, z * B.x - x * B.z, x * B.y - y * B.x);}
    inline double norm2() const {return x * x + y * y + z * z;}
    inline double norm() const {return sqrt(x * x + y * y + z * z);}
    inline bool operator < (const vec3 &B) const {return lt(x, B.x) || le(x, B.x) && (lt(y, B.y) || le(y, B.y) && lt(z, B.z));}
    inline bool operator == (const vec3 &B) const {return eq(x, B.x) && eq(y, B.y) && eq(z, B.z);}
};

// Positive if Right-hand rule
inline double volume(const vec3 A, const vec3 B, const vec3 C, const vec3 D) {return triple(B - A, C - A, D - A);}

struct line3 {
    vec3 P, t;
    line3 (vec3 _P = vec3(), vec3 _t = vec3()) : P(_P), t(_t) {}
};

inline double dis2(const vec3 P, const line3 l) {return ((P - l.P) ^ l.t).norm2() / l.t.norm2();}

struct plane {
    double A, B, C, D; // Ax + By + Cz + D = 0
    plane (double A0 = 0.0, double B0 = 0.0, double C0 = 0.0, double D0 = 0.0) : A(A0), B(B0), C(C0), D(D0) {}
    plane (const vec3 &u, const vec3 &v, const vec3 &w) {vec3 t = (v - u) ^ (w - u); A = t.x, B = t.y, C = t.z, D = -triple(u, v, w);} // > 0 if it follows Right-hand rule
    inline vec3 normVec() const {return vec3(A, B, C);}
    inline double norm2() const {return A * A + B * B + C * C;}
    inline double operator () (const vec3 &P) const {return A * P.x + B * P.y + C * P.z + D;}
};

inline double dis2(const vec3 P, const plane F) {return F(P) * F(P) / F.norm2();}

10 三维凸包 (卷包裹法)

namespace CH3D {
    typedef std::pair <int, int> pr;
    typedef std::set <pr> set;

    struct triangle {vec3 A, B, C;} tr[N];

    vec3 p[N], q[N], r[N];
    set E;
    int n, cnt;

    inline vec3 randvec3() {return vec3((double)rand() / RAND_MAX, (double)rand() / RAND_MAX, (double)rand() / RAND_MAX);}

    void wrap(int u, int v) {
        if (E.find({u, v}) == E.end()) {
            int i, w = -1;
            for (i = 0; i < n; ++i)
                if (i != u && i != v && (!~w || ge(volume(q[w], q[u], q[v], q[i]), 0))) w = i;
            if (~w) {
                tr[cnt++] = (triangle){p[u], p[v], p[w]};
                E.emplace(u, v); E.emplace(v, w); E.emplace(w, u);
                wrap(w, v); wrap(u, w);
            }
        }
    }

    void main(int _n, vec3 *_p) {
        int i;
        n = _n; cnt = 0; E.clear();
        memcpy(p, _p, n * sizeof(vec3));
        std::iter_swap(p, std::min_element(p, p + n));
        for (i = 0; i < n; ++i) q[i] = p[i] + randvec3() * 1e-6;
        for (i = 2; i < n; ++i)
            if (ge(cross(q[0], q[i], q[1]).z, 0)) std::swap(p[1], p[i]), std::swap(q[1], q[i]);
        wrap(0, 1);
    }
}

11 自适应 Simpson 积分

double Simpson(double L, double M, double R, double fL, double fM, double fR, double eps) {
    double LM = (L + M) * 0.5, fLM = f(LM), MR = (M + R) * 0.5, fMR = f(MR);
    double A = (fL + fM * 4.0 + fR) * (R - L) * sixth,
           AL = (fL + fLM * 4.0 + fM) * (M - L) * sixth,
           AR = (fM + fMR * 4.0 + fR) * (R - M) * sixth;
    if (fabs(AL + AR - A) < eps) return (2.0 * (AL + AR) + A) * third;
    return Simpson(L, LM, M, fL, fLM, fM, eps * 0.6) + Simpson(M, MR, R, fM, fMR, fR, eps * 0.6);
}