广州现场赛D题Signal Interference（计算几何）

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Signal Interference

题目链接

思路：推推公式就发现其实就是求一个圆和多边形面积的交

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);
}int n;
double xa, ya, xb, yb, k;int main() {int cas = 0;while (~scanf("%d%lf", &n, &k)) {for (int i = 0; i < n; i++) p[i].read();scanf("%lf%lf%lf%lf", &xa, &ya, &xb, &yb);Point o = Point(-(xb - k * k * xa) / (k * k - 1), -(yb - k * k * ya) / (k * k - 1));double r = (xb * xb + yb * yb - k * k * xa * xa - k * k * ya * ya) / (k * k - 1);r += (o.x * o.x + o.y * o.y);r = sqrt(r);printf("Case %d: %.10lf\n", ++cas, SPICA(n, o, r));}return 0;
}

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