POJ3675 Telescope(计算几何)

题意：求一个圆心在原点，半径r的圆和多边形的面积的交

思路：利用三角剖分，这题主要就是验证下模板

代码：

#include<stdio.h>#include<string.h>#include<stdlib.h>#include<math.h>#include<algorithm>const double eps = 1e-8;const double pi = acos(-1.0);int dcmp(double x){    if(x > eps) return 1;    return x < -eps ? -1 : 0;}struct Point{    double x, y;    Point(){x = y = 0;}    Point(double a, double b)    {x = a, y = b;}    inline void read()    {scanf("%lf%lf", &x, &y);}    inline Point operator-(const Point &b)const    {return Point(x - b.x, y - b.y);}    inline Point operator+(const Point &b)const    {return Point(x + b.x, y + b.y);}    inline Point operator*(const double &b)const    {return Point(x * b, y * b);}    inline double dot(const Point &b)const    {return x * b.x + y * b.y;}    inline double cross(const Point &b, const Point &c)const    {return (b.x - x) * (c.y - y) - (c.x - x) * (b.y - y);}    inline double Dis(const Point &b)const    {return sqrt((*this - b).dot(*this - b));}    inline bool InLine(const Point &b, const Point &c)const//三点共线    {return !dcmp(cross(b, c));}    inline bool OnSeg(const Point &b, const Point &c)const//点在线段上，包括端点    {return InLine(b, c) && (*this - c).dot(*this - b) < eps;}};inline double min(double a, double b){return a < b ? a : b;}inline double max(double a, double b){return a > b ? a : b;}inline double Sqr(double x){return x * x;}inline double Sqr(const Point &p){return p.dot(p);}Point LineCross(const Point &a, const Point &b, const Point &c, const Point &d){    double u = a.cross(b, c), v = b.cross(a, d);    return Point((c.x * v + d.x * u) / (u + v), (c.y * v + d.y * u) / (u + v));}double LineCrossCircle(const Point &a, const Point &b, const Point &r,            double R, Point &p1, Point &p2){    Point fp = LineCross(r, Point(r.x + a.y - b.y, r.y + b.x - a.x), a, b);    double rtol = r.Dis(fp);    double rtos = fp.OnSeg(a, b) ? rtol : min(r.Dis(a), r.Dis(b));    double atob = a.Dis(b);    double fptoe = sqrt(R * R - rtol * rtol) / atob;    if(rtos > R - eps) return rtos;    p1 = fp + (a - b) * fptoe;    p2 = fp + (b - a) * fptoe;    return rtos;}double SectorArea(const Point &r, const Point &a, const Point &b, double R)//不大于180度扇形面积，r->a->b逆时针{    double A2 = Sqr(r - a), B2 = Sqr(r - b), C2 = Sqr(a - b);    return R * R * acos((A2 + B2 - C2) * 0.5 / sqrt(A2) / sqrt(B2)) * 0.5;}double TACIA(const Point &r, const Point &a, const Point &b, double R)//TriangleAndCircleIntersectArea，逆时针，r为圆心{    double adis = r.Dis(a), bdis = r.Dis(b);    if(adis < R + eps && bdis < R + eps) return r.cross(a, b) * 0.5;    Point ta, tb;    if(r.InLine(a, b)) return 0.0;    double rtos = LineCrossCircle(a, b, r, R, ta, tb);    if(rtos > R - eps) return SectorArea(r, a, b, R);    if(adis < R + eps) return r.cross(a, tb) * 0.5 + SectorArea(r, tb, b, R);    if(bdis < R + eps) return r.cross(ta, b) * 0.5 + SectorArea(r, a, ta, R);    return r.cross(ta, tb) * 0.5 +        SectorArea(r, a, ta, R) + SectorArea(r, tb, b, R);}const int N = 505;Point p[N];double SPICA(int n, Point r, double R)//SimplePolygonIntersectCircleArea{    int i;    double res = 0, if_clock_t;    for(i = 0; i < n; ++ i)    {        if_clock_t = dcmp(r.cross(p[i], p[(i + 1) % n]));        if(if_clock_t < 0) res -= TACIA(r, p[(i + 1) % n], p[i], R);        else res += TACIA(r, p[i], p[(i + 1) % n], R);    }    return fabs(res);}double r;int n;int main() {    while (~scanf("%lf", &r)) {        scanf("%d", &n);        for (int i = 0; i < n; i++)            p[i].read();        printf("%.2f\n", SPICA(n, Point(0, 0), r));    }    return 0;}