CTU Open Contest 2019 C. Beer Coasters 计算几何 （计算几何模板）

C. Beer Coasters 计算几何

比赛地址：

https://www.jisuanke.com/contest/7321?view=challenges

题意：

给你一个圆和一个矩形，求两者相交的距离，其中矩形是以对角线形式给出。

基本思路：

计算几何，不会写，直接套多边形和圆相交模板，这里顺便存一个比较好用的计算几何模板。

#include 
#include 
#include 
#include 
#include 
using namespace std;

const double eps = 1e-8;
const double INF = 1e20;
const double pi = acos (-1.0);

int dcmp (double x) {
    if (fabs(x) < eps) return 0;
    return (x < 0 ? -1 : 1);
}
inline double sqr (double x) {return x*x;}

//*************点
struct Point {
    double x, y;
    Point(double _x = 0, double _y = 0) : x(_x), y(_y) {}
    void input() { scanf("%lf%lf", &x, &y); }
    void output() { printf("%.2f %.2f\n", x, y); }
    bool operator==(const Point &b) const {
        return (dcmp(x - b.x) == 0 && dcmp(y - b.y) == 0);
    }
    bool operator<(const Point &b) const {
        return (dcmp(x - b.x) == 0 ? dcmp(y - b.y) < 0 : x < b.x);
    }
    Point operator+(const Point &b) const {
        return Point(x + b.x, y + b.y);
    }
    Point operator-(const Point &b) const {
        return Point(x - b.x, y - b.y);
    }
    Point operator*(double a) {
        return Point(x * a, y * a);
    }
    Point operator/(double a) {
        return Point(x / a, y / a);
    }
    double len2() {//返回长度的平方
        return sqr(x) + sqr(y);
    }
    double len() {//返回长度
        return sqrt(len2());
    }
    Point change_len(double r) {//转化为长度为r的向量
        double l = len();
        if (dcmp(l) == 0) return *this;//零向量返回自身
        r /= l;
        return Point(x * r, y * r);
    }
    Point rotate_left() {//顺时针旋转90度
        return Point(-y, x);
    }
    Point rotate_right() {//逆时针旋转90度
        return Point(y, -x);
    }
    Point rotate(Point p, double ang) {//绕点p逆时针旋转ang
        Point v = (*this) - p;
        double c = cos(ang), s = sin(ang);
        return Point(p.x + v.x * c - v.y * s, p.y + v.x * s + v.y * c);
    }
    Point normal() {//单位法向量
        double l = len();
        return Point(-y / l, x / l);
    }
};

double cross (Point a, Point b) {//叉积
    return a.x * b.y - a.y * b.x;
}
double dot (Point a, Point b) {//点积
    return a.x * b.x + a.y * b.y;
}
double dis (Point a, Point b) {//两个点的距离
    Point p = b - a;
    return p.len();
}
double rad_degree (double rad) {//弧度转化为角度
    return rad / pi * 180;
}
double rad (Point a, Point b) {//两个向量的夹角
    return fabs(atan2(fabs(cross(a, b)), dot(a, b)));
}
bool parallel (Point a, Point b) {//向量平行
    double p = rad(a, b);
    return dcmp(p) == 0 || dcmp(p - pi) == 0;
}

//************直线 线段
struct Line {
    Point s, e;//直线的两个点
    Line() {}

    Line(Point _s, Point _e) : s(_s), e(_e) {}

    //一个点和倾斜角确定直线
    Line(Point p, double ang) {
        s = p;
        if (dcmp(ang - pi / 2) == 0) {
            e = s + Point(0, 1);
        } else
            e = s + Point(1, tan(ang));
    }

    //ax+by+c=0确定直线
    Line(double a, double b, double c) {
        if (dcmp(a) == 0) {
            s = Point(0, -c / b);
            e = Point(1, -c / b);
        } else if (dcmp(b) == 0) {
            s = Point(-c / a, 0);
            e = Point(-c / a, 1);
        } else {
            s = Point(0, -c / b);
            e = Point(1, (-c - a) / b);
        }
    }

    void input() {
        s.input();
        e.input();
    }

    void adjust() {
        if (e < s) swap(e, s);
    }

    double length() {//求线段长度
        return dis(s, e);
    }

    double angle() {//直线的倾斜角
        double k = atan2(e.y - s.y, e.x - s.x);
        if (dcmp(k) < 0) k += pi;
        if (dcmp(k - pi) == 0) k -= pi;
        return k;
    }
};

int relation (Point p, Line l) {//点和直线的关系
    //1:在左侧 2:在右侧 3:在直线上
    int c = dcmp(cross(p - l.s, l.e - l.s));
    if (c < 0) return 1;
    else if (c > 0) return 2;
    else return 3;
}

bool point_on_seg (Point p, Line l) {//判断点在线段上
    return dcmp(cross(p - l.s, l.e - l.s)) == 0 &&
           dcmp(dot(p - l.s, p - l.e) <= 0);
    //如果忽略端点交点改成小于号就好了
}

bool parallel (Line a, Line b) {//直线平行
    return parallel(a.e - a.s, b.e - b.s);
}

int seg_cross_seg (Line a, Line v) {//线段相交判断
    //1:规范相交 2:不规范相交 3:不相交
    int d1 = dcmp(cross(a.e - a.s, v.s - a.s));
    int d2 = dcmp(cross(a.e - a.s, v.e - a.s));
    int d3 = dcmp(cross(v.e - v.s, a.s - v.s));
    int d4 = dcmp(cross(v.e - v.s, a.e - v.s));
    if ((d1 ^ d2) == -2 && (d3 ^ d4) == -2) return 2;
    return (d1 == 0 && dcmp(dot(v.s - a.s, v.s - a.e)) <= 0) ||
           (d2 == 0 && dcmp(dot(v.e - a.s, v.e - a.e)) <= 0) ||
           (d3 == 0 && dcmp(dot(a.s - v.s, a.s - v.e)) <= 0) ||
           (d4 == 0 && dcmp(dot(a.e - v.s, a.e - v.e)) <= 0);
}

int line_cross_seg (Line a, Line v) {//直线和线段相交判断
    //1:规范相交 2:非规范相交 3:不相交
    int d1 = dcmp(cross(a.e - a.s, v.s - a.s));
    int d2 = dcmp(cross(a.e - a.s, v.e - a.s));
    if (d1 ^ d2 == -2) return 2;
    return (d1 == 0 || d2 == 0);
}

int line_cross_line (Line a, Line v) {//直线相交判断
    //0:平行 1:重合 2:相交
    if (parallel(a, v)) return relation(a.e, v) == 3;
    return 2;
}

Point line_intersection (Line a, Line v) {//直线交点
    //调用前确保有交点
    double a1 = cross(v.e - v.s, a.s - v.s);
    double a2 = cross(v.e - v.s, a.e - v.s);
    return Point((a.s.x * a2 - a.e.x * a1) / (a2 - a1), (a.s.y * a2 - a.e.y * a1) / (a2 - a1));
}

double point_to_line (Point p, Line a) {//点到直线的距离
    return fabs(cross(p - a.s, a.e - a.s) / a.length());
}

double point_to_seg (Point p, Line a) {//点到线段的距离
    if (dcmp(dot(p - a.s, a.e - a.s)) < 0 || dcmp(dot(p - a.e, a.s - a.e)) < 0)
        return min(dis(p, a.e), dis(p, a.s));
    return point_to_line(p, a);
}

Point projection (Point p, Line a) {//点在直线上的投影
    return a.s + (((a.e - a.s) * dot(a.e - a.s, p - a.s)) / (a.e - a.s).len2());
}

Point symmetry (Point p, Line a) {//点关于直线的对称点
    Point q = projection(p, a);
    return Point(2 * q.x - p.x, 2 * q.y - p.y);
}

//***************圆
struct Circle {
    //圆心 半径
    Point p;
    double r;
    Circle() {}
    Circle(Point _p, double _r) : p(_p), r(_r) {}
    Circle(double a, double b, double _r) {
        p = Point(a, b);
        r = _r;
    }
    void input() {
        p.input();
        scanf("%lf", &r);
    }
    void output() {
        p.output();
        printf(" %.2f\n", r);
    }
    bool operator==(const Circle &a) const {
        return p == a.p && (dcmp(r - a.r) == 0);
    }
    double area() {//面积
        return pi * r * r;
    }
    double circumference() {//周长
        return 2 * pi * r;
    }
};

int relation (Point p, Circle a) {//点和圆的关系
    //0:圆外 1:圆上 2:圆内
    double d = dis(p, a.p);
    if (dcmp(d - a.r) == 0) return 1;
    return (dcmp(d - a.r) < 0 ? 2 : 0);
}

int relation (Line a, Circle b) {//直线和圆的关系
    //0:相离 1:相切 2:相交
    double p = point_to_line(b.p, a);
    if (dcmp(p - b.r) == 0) return 1;
    return (dcmp(p - b.r) < 0 ? 2 : 0);
}

int relation (Circle a, Circle v) {//两圆的位置关系
    //1:内含 2:内切 3:相交 4:外切 5:相离
    double d = dis(a.p, v.p);
    if (dcmp(d - a.r - v.r) > 0) return 5;
    if (dcmp(d - a.r - v.r) == 0) return 4;
    double l = fabs(a.r - v.r);
    if (dcmp(d - a.r - v.r) < 0 && dcmp(d - l) > 0) return 3;
    if (dcmp(d - l) == 0) return 2;
    if (dcmp(d - l) < 0) return 1;
}

Circle out_circle (Point a, Point b, Point c) {//三角形外接圆
    Line u = Line((a + b) / 2, ((a + b) / 2) + (b - a).rotate_left());
    Line v = Line((b + c) / 2, ((b + c) / 2) + (c - b).rotate_left());
    Point p = line_intersection(u, v);
    double r = dis(p, a);
    return Circle(p, r);
}

Circle in_circle (Point a, Point b, Point c) {//三角形内切圆
    Line u, v;
    double m = atan2(b.y - a.y, b.x - a.x), n = atan2(c.y - a.y, c.x - a.x);
    u.s = a;
    u.e = u.s + Point(cos((n + m) / 2), sin((n + m) / 2));
    v.s = b;
    m = atan2(a.y - b.y, a.x - b.x), n = atan2(c.y - b.y, c.x - b.x);
    v.e = v.s + Point(cos((n + m) / 2), sin((n + m) / 2));
    Point p = line_intersection(u, v);
    double r = point_to_seg(p, Line(a, b));
    return Circle(p, r);
}

int circle_intersection (Circle a, Circle v, Point &p1, Point &p2) {//两个圆的交点
    //返回交点个数 交点保存在引用中
    int rel = relation(a, v);
    if (rel == 1 || rel == 5) return 0;
    double d = dis(a.p, v.p);
    double l = (d * d + a.r * a.r - v.r * v.r) / (2 * d);
    double h = sqrt(a.r * a.
            r - l * l);
    Point tmp = a.p + (v.p - a.p).change_len(l);
    p1 = tmp + ((v.p - a.p).rotate_left().change_len(h));
    p2 = tmp + ((v.p - a.p).rotate_right().change_len(h));
    if (rel == 2 || rel == 4) return 1;
    return 2;
}

int line_cirlce_intersection (Line v, Circle u, Point &p1, Point &p2) {//直线和圆的交点
    //返回交点个数 交点保存在引用中
    if (!relation(v, u)) return 0;
    Point a = projection(u.p, v);
    double d = point_to_line(u.p, v);
    d = sqrt(u.r * u.r - d * d);
    if (dcmp(d) == 0) {
        p1 = a, p2 = a;
        return 1;
    }
    p1 = a + (v.e - v.s).change_len(d);
    p2 = a - (v.e - v.s).change_len(d);
    return 2;
}

int get_circle (Point a, Point b, double r1, Circle &c1, Circle &c2) {//得到过ab半径为r1的了两个圆
    //返回得到圆的个数 圆保存在两个引用中
    Circle x(a, r1), y(b, r1);
    int t = circle_intersection(x, y, c1.p, c2.p);
    if (!t) return 0;
    c1.r = c2.r = r1;
    return t;
}

int get_circle (Line u, Point q, double r1, Circle &c1, Circle &c2) {//得到和直线u相切 过点q 半径为r1的圆
    double d = point_to_line(q, u);
    if (dcmp(d - r1 * 2) > 0) return 0;
    if (dcmp(d) == 0) {
        c1.p = q + ((u.e - u.s).rotate_left().change_len(r1));
        c2.p = q + ((u.e - u.s).rotate_right().change_len(r1));
        c1.r = c2.r = r1;
        return 2;
    }
    Line u1 = Line(u.s + (u.e - u.s).rotate_left().change_len(r1), u.e + (u.e - u.s).rotate_left().change_len(r1));
    Line u2 = Line(u.s + (u.e - u.s).rotate_right().change_len(r1), u.e + (u.e - u.s).rotate_right().change_len(r1));
    Circle cc = Circle(q, r1);
    Point p1, p2;
    if (!line_cirlce_intersection(u1, cc, p1, p2))
        line_cirlce_intersection(u2, cc, p1, p2);
    c1 = Circle(p1, r1);
    if (p1 == p2) {
        c2 = c1;
        return 1;
    }
    c2 = Circle(p2, r1);
    return 2;
}

int get_circle (Line u, Line v, double r1, Circle &c1, Circle &c2, Circle &c3, Circle &c4) {//和直线u,v相切 半径为r1的圆
    if (parallel(u, v)) return 0;
    Line u1 = Line(u.s + (u.e - u.s).rotate_left().change_len(r1), u.e + (u.e - u.s).rotate_left().change_len(r1));
    Line u2 = Line(u.s + (u.e - u.s).rotate_right().change_len(r1), u.e + (u.e - u.s).rotate_right().change_len(r1));
    Line v1 = Line(v.s + (v.e - v.s).rotate_left().change_len(r1), v.e + (v.e - v.s).rotate_left().change_len(r1));
    Line v2 = Line(v.s + (v.e - v.s).rotate_right().change_len(r1), v.e + (v.e - v.s).rotate_right().change_len(r1));
    c1.r = c2.r = c3.r = c4.r = r1;
    c1.p = line_intersection(u1, v1);
    c2.p = line_intersection(u1, v2);
    c3.p = line_intersection(u2, v1);
    c4.p = line_intersection(u2, v2);
    return 4;
}

int get_circle (Circle cx, Circle cy, double r1, Circle &c1, Circle &c2) {//和两个圆相切 半径为r1的圆
    //确保两个圆外离
    Circle x(cx.p, r1 + cx.r), y(cy.p, r1 + cy.r);
    int t = circle_intersection(x, y, c1.p, c2.p);
    if (!t) return 0;
    c1.r = c2.r = r1;
    return t;
}

int tan_line (Point q, Circle a, Line &u, Line &v) {//过一点作圆切线
    int x = relation(q, a);
    if (x == 2) return 0;
    if (x == 1) {
        u = Line(q, q + (q - a.p).rotate_left());
        v = u;
        return 1;
    }
    double d = dis(a.p, q);
    double l = a.r * a.r / d;
    double h = sqrt(a.r * a.r - l * l);
    u = Line(q, a.p + (q - a.p).change_len(l) + (q - a.p).rotate_left().change_len(h));
    v = Line(q, a.p + (q - a.p).change_len(l) + (q - a.p).rotate_right().change_len(h));
    return 2;
}

double area_circle (Circle a, Circle v) {//两圆相交面积
    int rel = relation(a, v);
    if (rel >= 4) return 0;
    if (rel <= 2) return min(a.area(), v.area());
    double d = dis(a.p, v.p);
    double hf = (a.r + v.r + d) / 2;
    double ss = 2 * sqrt(hf * (hf - a.r) * (hf - v.r) * (hf - d));
    double a1 = acos((a.r * a.r + d * d - v.r * v.r) / (2 * a.r * d));
    a1 = a1 * a.r * a.r;
    double a2 = acos((v.r * v.r + d * d - a.r * a.r) / (2 * v.r * d));
    a2 = a2 * v.r * v.r;
    return a1 + a2 - ss;
}

double circle_traingle_area (Point a, Point b, Circle c) {//圆心三角形的面积
    //a.output (), b.output (), c.output ();
    Point p = c.p;
    double r = c.r; //cout << cross (p-a, p-b) << endl;
    if (dcmp(cross(p - a, p - b)) == 0) return 0;
    Point q[5];
    int len = 0;
    q[len++] = a;
    Line l(a, b);
    Point p1, p2;
    if (line_cirlce_intersection(l, c, q[1], q[2]) == 2) {
        if (dcmp(dot(a - q[1], b - q[1])) < 0) q[len++] = q[1];
        if (dcmp(dot(a - q[2], b - q[2])) < 0) q[len++] = q[2];
    }
    q[len++] = b;
    if (len == 4 && dcmp(dot(q[0] - q[1], q[2] - q[1])) > 0)
        swap(q[1], q[2]);
    double res = 0;
    for (int i = 0; i < len - 1; i++) {
        if (relation(q[i], c) == 0 || relation(q[i + 1], c) == 0) {
            double arg = rad(q[i] - p, q[i + 1] - p);
            res += r * r * arg / 2.0;
        } else {
            res += fabs(cross(q[i] - p, q[i + 1] - p)) / 2;
        }
    } //cout << res << ".." << endl;
    return res;
}

//*************多边形
bool is_convex (Point *p, int n) {//判断n边行是不是凸的
    bool s[2];
    s[0] = s[1] = 0;
    for (int i = 0; i < n; i++) {
        int j = (i + 1) % n;
        int k = (j + 1) % n;
        s[dcmp(cross(p[j] - p[i], p[k] - p[i])) + 1] = 1;
        if (s[0] && s[2]) return 0;
    }
    return 1;
}

double polygon_area (Point *p, int n) {//多边形的有向面积,加上绝对值就是面积
    //n个点
    double area = 0;
    for (int i = 1; i < n - 1; i++) {
        area += cross(p[i] - p[0], p[i + 1] - p[0]);
    }
    return area / 2;
}

bool relation (Point a, Point *b, int n) {//点和多边形的关系(凸凹都可以)
    //0:外部 1:内部 2:边上 3:顶点
    int w = 0;
    for (int i = 0; i < n; i++) {
        if (a == b[i] || a == b[(i + 1) % n])
            return 3;
        if (point_on_seg(a, Line(b[(i + 1) % n], b[i])))
            return 2;
        int k = dcmp(cross(b[(i + 1) % n] - b[i], a - b[i]));
        int d1 = dcmp(b[i].y - a.y);
        int d2 = dcmp(b[(i + 1) % n].y - a.y);
        if (k > 0 && d1 <= 0 && d2 > 0)
            w++;
        if (k < 0 && d2 <= 0 && d1 > 0)
            w--;
    }
    if (w != 0)
        return 1;
    return 0;
}

int convex_cut (Line u, Point *p, int n, Point *po) {//直线切割多边形左侧
    //返回切割后多边形的数量
    int top = 0;
    for (int i = 0; i < n; i++) {
        int d1 = dcmp(cross(u.e - u.s, p[i] - u.s));
        int d2 = dcmp(cross(u.e - u.s, p[(i + 1) % n] - u.s));
        if (d1 >= 0) po[top++] = p[i];
        if (d1 * d2 < 0) po[top++] = line_intersection(u, Line(p[i], p[(i + 1) % n]));
    }
    return top;
}

double convex_circumference (Point *p, int n) {//多边形的周长(凹凸都可以)
    double ans = 0;
    for (int i = 0; i < n; i++) {
        ans += dis(p[i], p[(i + 1) % n]);
    }
    return ans;
}

double area_polygon_circle (Circle c, Point *p, int n) {//多边形和圆交面积
    double ans = 0;
    for (int i = 0; i < n; i++) {
        int j = (i + 1) % n; //cout << i << " " << j << "//" << endl;
        if (dcmp(cross(p[j] - c.p, p[i] - c.p)) >= 0)
            ans += circle_traingle_area(p[i], p[j], c);
        else
            ans -= circle_traingle_area(p[i], p[j], c);
    }
    return fabs(ans);
}

Point centre_of_gravity (Point *p, int n) {//多边形的重心(凹凸都可以)
    double sum = 0.0, sumx = 0, sumy = 0;
    Point p1 = p[0], p2 = p[1], p3;
    for (int i = 2; i <= n - 1; i++) {
        p3 = p[i];
        double area = cross(p2 - p1, p3 - p2) / 2.0;
        sum += area;
        sumx += (p1.x + p2.x + p3.x) * area;
        sumy += (p1.y + p2.y + p3.y) * area;
        p2 = p3;
    }
    return Point(sumx / (3.0 * sum), sumy / (3.0 * sum));
}

int convex_hull (Point *p, Point *ch, int n) {//求凸包
    //所有的点集 凸包点集 点集的点数
    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 - 1]) <= 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 - 1]) <= 0)
            m--;
        ch[m++] = p[i];
    }
    if (n > 1)
        m--;
    return m;
}

const int maxn = 10;
Point p[maxn];

int main () {
    double x,y,r;
    cin >> x >> y >> r;
    Circle o = Circle(Point(x,y),r);
    double a1,a2,b1,b2;
    cin >> a1 >> a2 >> b1 >> b2;
    //点的录入要保证顺逆时针关系;
    p[0] = Point(a1,a2),p[1] = Point(a1,b2),p[2] = Point(b1,b2),p[3] = Point(b1,a2);
    double ans = area_polygon_circle(o,p,4);
    printf("%.4lf\n",ans);
    return 0;
}