圓形與矩形截面的面積

三角仍然可以做到這一點

代碼：

#include<stdio.h> #include<string.h> #include<stdlib.h> #include<math.h> #include<algorithm> using namespace std;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//三點共線{return !dcmp(cross(b, c));}inline bool OnSeg(const Point &b, const Point &c)const//點在線段上，包含端點{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度扇形面積。r->a->b逆時針 { 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。逆時針，r為圓心 { 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], o; 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 main() { int bo = 0; while (~scanf("%lf%lf%lf", &o.x, &o.y, &r)) { if (bo) printf("\n"); else bo = 1; double x1, y1, x2, y2; scanf("%lf%lf%lf%lf", &x1, &y1, &x2, &y2); if (x1 > x2) swap(x1, x2); if (y1 > y2) swap(y1, y2); p[0] = Point(x1, y1); p[1] = Point(x1, y2); p[2] = Point(x2, y2); p[3] = Point(x2, y1); printf("%.10f\n", SPICA(4, o, r)); } return 0; }

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轉載于:https://www.cnblogs.com/mengfanrong/p/4752774.html

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