HDU 3982 (半平呈送 多边形和圆面积交)

题目链接：点击这里

题意：一块圆形蛋糕， 上面有一个点是樱桃， 然后切n刀， 求樱桃所在的那一块面积占总面积的百分比。

n刀每次取樱桃所在的半平面， 然后最外面加一个大矩形限制半平面交，最后求出的半平面交和圆面积并一下就是樱桃那块的面积了。

#include <cstdio>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <vector>
using namespace std;
#define maxn 100005

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
    double k;//极角
    Line () {}
    Line (Point _s, Point _e) {
        s = _s, e = _e;
        k = atan2 (e.y - s.y,e.x - s.x);
    }
    //ax+by+c = 0
    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);
        }
        get_angle ();
    }
    //一个点和倾斜角确定直线
    Line (Point p, double ang) {
        k = ang;
        s = p;
        if (dcmp (ang-pi/2) == 0) {
            e = s + Point (0, 1);
        }
        else
            e = s + Point (1, tan (ang));
    }
    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);
    }
    void get_angle () {
        k = atan2 (e.y - s.y,e.x - s.x);
    }
    double angle () {//直线的倾斜角
        if (dcmp (k) < 0) k += pi;
        if (dcmp (k-pi) == 0) k -= pi;
        return k;
    }
    Point operator &(const Line &b)const {//直线的交点(保证存在)
        Point res = s;
        double t = (cross (s - b.s, b.s - b.e))/cross (s - e, b.s - b.e); 
        res.x += (e.x - s.x)*t;
        res.y += (e.y - s.y)*t;
        return res;
    }
};

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;
}

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 ());
}

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() );
}

//***************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;
    }
    bool operator < (const Circle &a) const {
        return p < a.p || (p == a.p && r < a.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 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;
}

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;
}

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; 
        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);
}

//半平面交,直线的左边代表有效区域 
bool HPIcmp (const Line &a, const Line &b) {
    if (fabs(a.k - b.k) > eps)
        return a.k < b.k;
    return cross (a.s - b.s, b.e - b.s) < 0; 
}
Line Q[maxn];
void HPI(Line line[], int n, Point res[], int &resn) {
    //半平面数组 半平面个数 半平面交顶点 半平面交顶点个数
    for (int i = 0; i < n; i++) line[i].get_angle ();
    int tot = n; 
    sort(line,line+n,HPIcmp); 
    tot = 1;
    for(int i = 1;i < n;i++){
        if(fabs(line[i].k - line[i-1].k) > eps)
            line[tot++] = line[i];
    }
    int head = 0, tail = 1; 
    Q[0] = line[0];
    Q[1] = line[1];
    resn = 0;
    for (int i = 2; i < tot; i++) {
        if (fabs(cross (Q[tail].e-Q[tail].s, Q[tail-1].e-Q[tail-1].s)) < eps || fabs(cross (Q[head].e-Q[head].s, Q[head+1].e-Q[head+1].s)) < eps)
            return;
        while(head < tail && (cross ((Q[tail]&Q[tail-1])-line[i].s, line[i].e-line[i].s)) > eps) tail--;
        while(head < tail && (cross ((Q[head]&Q[head+1]) - line[i].s, line[i].e-line[i].s)) > eps)
        head++; 
        Q[++tail] = line[i];
    }
    while(head < tail && (cross ((Q[tail]&Q[tail-1]) - Q[head].s, Q[head].e-Q[head].s)) > eps)
        tail--;
    while(head < tail && (cross ((Q[head]&Q[head-1]) -Q[tail].s, Q[tail].e-Q[tail].e)) > eps) 
        head++;
    if(tail <= head + 1)
        return; 
    for(int i = head; i < tail; i++)
        res[resn++] = Q[i]&Q[i+1]; 
    if(head < tail - 1)
        res[resn++] = Q[head]&Q[tail];
}

double r;
int n, m;
Line l[maxn];
Point O, ans[maxn];
Line hp[maxn];

int main () {
    int t, kase = 0;
    scanf ("%d", &t);
    while (t--) {
        scanf ("%lf%d", &r, &n);
        for (int i = 0; i < n; i++) {
            l[i].input ();
        }
        O.input ();
        int cnt = 0;
        hp[cnt++] = Line (Point (r, -r), Point (r, r));
        hp[cnt++] = Line (Point (r, r), Point (-r, r));
        hp[cnt++] = Line (Point (-r, r), Point (-r, -r));
        hp[cnt++] = Line (Point (-r, -r), Point (r, -r));
        for (int i = 0; i < n; i++) {
            if (relation (O, l[i]) == 1) {
                hp[cnt++] = l[i];
            }
            else 
                hp[cnt++] = Line (l[i].e, l[i].s);
        }
        HPI (hp, cnt, ans, m);
        Circle C (Point (0, 0), r);
        double area = area_polygon_circle (C, ans, m);
        double tot = pi*r*r;
        area = area * 100 / tot;
        printf ("Case %d: %.5f%%\n", ++kase, area);
    }
    return 0;
}