题意：略
思路：多边形与圆相交我记得好像是转化成圆与三角形相交，然后再求；
代码：
#include <bits/stdc++.h> using namespace std; const int maxn = 110; const double PI = acos(-1); const double TWO_PI = PI * 2; const double eps=1e-8; double NormalizeAngle(double rad, double center = PI) { return rad - TWO_PI * floor((rad + PI - center) / TWO_PI); } struct Point { double x, y; Point (double x=0, double y=0):x(x), y(y) {} double operator ^(const Point &b)const { return x*b.y - y*b.x; } }; typedef Point Vector; 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); } }; struct Line { Point p; Vector v; double ang; Line(){} Line(Point p, Vector v) : p(p), v(v) {ang = atan2(v.y, v.x); } Point point(double t) { return Point(p.x + t*v.x, p.y + t*v.y); } bool operator < (const Line& L) const { return ang < L.ang; } }; template <class T> T sqr(T x) { return x * x;} Vector operator + (Vector A, Vector B) { return Vector(A.x + B.x, A.y + B.y); } Vector operator - (Vector A, Vector B) { return Vector(A.x - B.x, A.y - B.y); } Vector operator * (Vector A, double p) { return Vector(A.x*p, A.y*p); } Vector operator / (Vector A, double p) { return Vector(A.x/p, A.y/p); } bool operator < (const Point& a, const Point& b) { return a.x < b.x || (a.x == b.x && a.y < b.y); } int dcmp(double x) { if(fabs(x) < eps) return 0; return x < 0 ? -1 : 1; } bool operator == (const Point& a, const Point& b){ return dcmp(a.x-b.x) == 0 && dcmp(a.y-b.y) == 0;} double Dot(Vector A, Vector B) { return A.x*B.x + A.y*B.y; } double Length(Vector A) { return sqrt(Dot(A, A)); } double Angel(Vector A, Vector B) { return acos(Dot(A, B) / Length(A) / Length(B)); } double Cross(Vector A, Vector B) { return A.x*B.y - A.y*B.x; } double torad(double d) { return (d/180)*PI; } Vector vecunit(Vector x){ return x / Length(x);} //单位向量 Vector normal(Vector x) { return Point(-x.y, x.x) / Length(x);} //垂直法向量 Point GetIntersection(Line a, Line b) //线段交点 { Vector u = a.p-b.p; double t = Cross(b.v, u) / Cross(a.v, b.v); return a.p + a.v*t; } bool OnSegment(Point p, Point a1, Point a2) { return dcmp(Cross(a1-p, a2-p)) == 0 && dcmp(Dot(a1-p, a2-p)) < 0; } bool InCircle(Point x, Circle c) { return dcmp(sqr(c.r) - sqr(Length(c.c - x))) >= 0;} bool OnCircle(Point x, Circle c) { return dcmp(sqr(c.r) - sqr(Length(c.c - x))) == 0;} double angle(Vector x) { return atan2(x.y, x.x);} //直线与圆交点 int getLineCircleIntersection(Line L, Circle C, double& t1, double& t2, vector<Point>& sol) { 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); sol.push_back(L.point(t1)); return 1; } t1 = (-f-sqrt(delta))/(2*e); sol.push_back(L.point(t1)); t2 = (-f+sqrt(delta))/(2*e); sol.push_back(L.point(t2)); return 2; } //线段与圆的焦点 int getSegCircleIntersection(Line L, Circle C, Point* sol) { Vector nor = normal(L.v); Line pl = Line(C.c, nor); Point ip = GetIntersection(pl, L); double dis = Length(ip - C.c); if (dcmp(dis - C.r) > 0) return 0; Point dxy = vecunit(L.v) * sqrt(sqr(C.r) - sqr(dis)); int ret = 0; sol[ret] = ip + dxy; if (OnSegment(sol[ret], L.p, L.point(1))) ret++; sol[ret] = ip - dxy; if (OnSegment(sol[ret], L.p, L.point(1))) ret++; return ret; } double SegCircleArea(Circle C, Point a, Point b) //线段切割圆 { double a1 = angle(a - C.c); double a2 = angle(b - C.c); double da = fabs(a1 - a2); if (da > PI) da = PI * 2.0 - da; return dcmp(Cross(b - C.c, a - C.c)) * da * sqr(C.r) / 2.0; } double PolyCiclrArea(Circle C, Point *p, int n)//多边形与圆相交面积 { double ret = 0.0; Point sol[2]; p[n] = p[0]; for(int i=0;i<n;i++) { double t1, t2; int cnt = getSegCircleIntersection(Line(p[i], p[i+1]-p[i]), C, sol); if (cnt == 0) { if (!InCircle(p[i], C) || !InCircle(p[i+1], C)) ret += SegCircleArea(C, p[i], p[i+1]); else ret += Cross(p[i+1] - C.c, p[i] - C.c) / 2.0; } if (cnt == 1) { if (InCircle(p[i], C) && !InCircle(p[i+1], C)) ret += Cross(sol[0] - C.c, p[i] - C.c) / 2.0, ret += SegCircleArea(C, sol[0], p[i+1]); else ret += SegCircleArea(C, p[i], sol[0]), ret += Cross(p[i+1] - C.c, sol[0] - C.c) / 2.0; } if (cnt == 2) { if ((p[i] < p[i + 1]) ^ (sol[0] < sol[1])) swap(sol[0], sol[1]); ret += SegCircleArea(C, p[i], sol[0]); ret += Cross(sol[1] - C.c, sol[0] - C.c) / 2.0; ret += SegCircleArea(C, sol[1], p[i+1]); } } return fabs(ret); } double CalcArea(Point p[],int n) { double res = 0; for(int i = 0; i < n; i++) res += (p[i]^p[(i+1)%n])/2; return fabs(res); } int n; double x, y, v, ang, t, g, r; Circle C; Point p[maxn]; int main() { while (scanf("%lf%lf%lf%lf%lf%lf%lf", &x, &y, &v, &ang, &t, &g, &r)){ if(x==0&&y==0&&v==0&&ang==0&&t==0&&g==0&&r==0) break; ang=torad(ang); C=Circle(Point(x + v*cos(ang)*t, y + v*sin(ang)*t - 0.5*g*t*t), r); scanf("%d", &n); for(int i=0;i<n;i++)scanf("%lf%lf", &p[i].x, &p[i].y); printf("%.2f ", PolyCiclrArea(C, p, n)); } return 0; }