题意：一个多边形，A点和B点，满足PB <= k * PA的P的范围与多边形的公共面积。

分析：这是个。既然是圆，那么设圆的一般方程：(x + D/2) ^ 2 + (y + E/2) ^ 2 = (D ^ 2 + E ^ 2 - 4 * F)  / 4，通过PB == PA * k解方程来求解圆心以及半径。然后就是套模板啦，上海交大的红书。

 

#include 
     
      using namespace std; #define lson l, mid, rt << 1#define rson mid + 1, r, rt << 1 | 1typedef long long ll;const int N = 5e2 + 10;const int INF = 0x3f3f3f3f;const int MOD = 1e9 + 7;const double EPS = 1e-10;const double PI = acos (-1.0);int dcmp(double x)  {    if (fabs (x) < EPS)   return 0;    else    return x < 0 ? -1 : 1;}struct Point    {    double x, y;    Point ()    {}    Point (double x, double y) : x (x), y (y) {}    Point operator + (const Point &r) const {       //向量加法        return Point (x + r.x, y + r.y);    }    Point operator - (const Point &r) const {        return Point (x - r.x, y - r.y);    }    Point operator * (double p) const {       //向量乘以标量        return Point (x * p, y * p);    }    Point operator / (double p) const {       //向量除以标量        return Point (x / p, y / p);    }    bool operator < (const Point &r) const  {        return x < r.x || (x == r.x && y < r.y);    }    bool operator == (const Point &r) const {        return dcmp (x - r.x) == 0 && dcmp (y - r.y) == 0;    }};typedef Point Vector;Point read_point(void)  {    double x, y;    scanf ("%lf%lf", &x, &y);    return Point (x, y);}double polar_angle(Vector V)    {    return atan2 (V.y, V.x);}double dot(Point a, Point b)    {    return a.x * b.x + a.y * b.y;}double cross(Point a, Point b)  {    return a.x * b.y - a.y * b.x;}double length(Vector V) {    return sqrt (dot (V, V));}double my_sqrt(double x)    {    return sqrt (max (0.0, x));} Point ps[N];double r, k;double x1, _y1, x2, _y2;int n; double sqr(double x)    {    return x * x;} struct  Line    {    Point p;    Vector v;    double r;    Line () {}    Line (const Point &p, const Vector &v) : p (p), v (v) {        r = polar_angle (v);    }    Point point(double a)   {        return p + v * a;    }}; struct Circle   {    Point c;    double r;    Circle () {}    Circle (Point c, double r) : c (c), r (r) {}    Point point(double a)   {        return Point (c.x + cos (a) * r, c.y + sin (a) * r);    }}; int line_cir_inter(Line L, Circle C, double &t1, double &t2, vector
      
        &P)    {    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;    if (dcmp (delta) < 0)   return 0;    if (dcmp (delta) == 0)  {        t1 = t2 = -f / (2 * e); P.push_back (L.point (t1));        return 1;    }    t1 = (-f - sqrt (delta)) / (2 * e);    t2 = (-f + sqrt (delta)) / (2 * e);    if (t1 > t2)    swap (t1, t2);    if (dcmp (t1) > 0 && dcmp (t1 - 1) < 0) P.push_back (L.point (t1));    if (dcmp (t2) > 0 && dcmp (t2 - 1) < 0) P.push_back (L.point (t2));    return (int) P.size ();} double sector_area(Point a, Point b)    {    double theta = polar_angle (a) - polar_angle (b);    while (dcmp (theta) <= 0)   theta += 2 * PI;    while (theta > 2 * PI)  theta -= 2 * PI;    theta = min (theta, 2 * PI - theta);    return r * r * theta / 2;} double cal(Point a, Point b)    {    double t1, t2;    bool ina = dcmp (length (a) - r) < 0;    bool inb = dcmp (length (b) - r) < 0;    if (ina && inb) return fabs (cross (a, b)) / 2.0;    vector
       
         p;    int num = line_cir_inter (Line (a, b - a), Circle (Point (0, 0), r), t1, t2, p);    if (ina)    return sector_area (b, p[0]) + fabs (cross (a, p[0])) / 2.0;    if (inb)    return sector_area (p[0], a) + fabs (cross (p[0], b)) / 2.0;    if (num == 2)   return sector_area (a, p[0]) + sector_area (p[1], b) + fabs (cross (p[0], p[1])) / 2.0;    return sector_area (a, b);} double cir_poly_area()  {    double ret = 0;    for (int i=0; i