题意：按顺时针或逆时针顺序给出一个凸n边形的n个点的坐标，然后让一个圆心在(0,0)的圆和凸n边形相交的面积大于等于R，问圆的最小半径。
题解：这题简直坑爹啊，各种细节错误。。修修改改了一天，最后看别人题解也还是不懂为什么OnSegment函数要写成那样。。。明明不能判断点是否在线段上 ?(???)?
画画图思路不难想到，把凸n边形的每条边都和圆判断关系，如果是边的两点都在圆内，两条边对应一个三角形的面积，如果一个点在圆外一个在圆内(包括边界)，那就是一个三角形加一个扇形，如果两点都在圆外，分两种情况，和圆有两个交点的话是两个扇形和一个三角形，如果和圆无交点是一个扇形。

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         using namespace std; const double eps = 1e-9; const double PI = acos(-1); int dcmp(double x) { if (fabs(x) < eps) return 0; return x > 0 ? 1 : -1; } struct Point { double x, y; Point (double a = 0, double b = 0): x(a), y(b) {} }; typedef Point Vector; Vector operator + (const Vector& a, const Vector& b) { return Vector(a.x + b.x, a.y + b.y); } Vector operator - (const Vector& a, const Vector& b) { return Vector(a.x - b.x, a.y - b.y); } Vector operator * (const Vector& a, double& b) { return Vector(a.x * b, a.y * b); } Vector operator / (const Vector& a, double& b) { return Vector(a.x / b, a.y / b); } bool operator == (const Vector& a, const Vector& b) { return !dcmp(a.x - b.x) && !dcmp(a.y - b.y); } bool operator < (const Vector& a, const Vector& b) { return a.x < b.x || (a.x == b.x && a.y < b.y); } double Dot(const Vector& a, const Vector& b) { return a.x * b.x + a.y * b.y; } double Length(const Vector& a) { return sqrt(Dot(a, a)); } double Cross(const Vector& a, const Vector& b) { return a.x * b.y - a.y * b.x; } double Angle(const Vector& a, const Vector& b) { return acos(Dot(a, b) / Length(a) / Length(b)); } struct Line { Point p; Vector v; double ang; Line() {} Line(Point a, Vector b): p(a), v(b) { ang = atan2(b.y, b.x); } bool operator < (const Line& L) const { return ang < L.ang; } Point point(double a) { return p + v * a; } }; struct Circle { double r; Point c; Circle() {} Circle(Point a, double b): c(a), r(b) {} Point point(double a) { return Point(c.x + cos(a) * r, c.y + sin(a) * r); } }; bool isPointInCircle(Point P, Circle C) { return dcmp(Length(C.c - P) - C.r) <= 0; } bool OnSegment(Point p, Point p1, Point p2) { return dcmp(Dot(p1 - p, p2 - p)) <= 0;//这里很费解 } void SegmentCircleIntersection(Point p1, Point p2, Circle C, Point* sol, int& num) { double t1, t2; double a = (p1 - C.c).x, b = (p2 - p1).x, c = (p1 - C.c).y, d = (p2 - p1).y; double e = b * b + d * d, f = 2 * (a * b + c * d), g = a * a + c * c - C.r * C.r; double delta = f * f - 4 * e * g; Point p; num = 0; if (dcmp(delta) < 0) return; else if (dcmp(delta) == 0) { t1 = t2 = (-f) / (2 * e); p = p1 + (p2 - p1) * t1; if (OnSegment(p, p1, p2)) sol[num++] = p; } else { t1 = (-f - sqrt(delta)) / (2 * e); p = p1 + (p2 - p1) * t1; if (OnSegment(p, p1, p2)) sol[num++] = p; t2 = (-f + sqrt(delta)) / (2 * e); p = p1 + (p2 - p1) * t2; if (OnSegment(p, p1, p2)) sol[num++] = p; } } double FanArea(Circle C, Vector v1, Vector v2) { double rad = Angle(v1, v2); if (dcmp(rad - PI) == 0) rad = 0; return C.r * C.r * rad * 0.5; } double GetCommonArea(Point p1, Point p2, Point p3, Circle C) { int flag1 = isPointInCircle(p2, C); int flag2 = isPointInCircle(p3, C); int num; Point sol[5]; if (flag1 + flag2 == 2) return fabs(Cross(p2 - p1, p3 - p1)) * 0.5; if (flag1 + flag2 == 1) { SegmentCircleIntersection(p2, p3, C, sol, num); if (flag1) return FanArea(C, p3 - p1, sol[0] - p1) + fabs(Cross(p2 - p1, sol[0] - p1)) * 0.5; return FanArea(C, p2 - p1, sol[0] - p1) + fabs(Cross(p3 - p1, sol[0] - p1)) * 0.5; } SegmentCircleIntersection(p2, p3, C, sol, num);//注意两个端点对应不同交点，写反就错了 if (num == 2) return FanArea(C, p2 - p1, sol[0] - p1) + fabs(Cross(sol[0] - p1, sol[1] - p1)) * 0.5 + FanArea(C, p3 - p1, sol[1] - p1); return FanArea(C, p2 - p1, p3 - p1); } const int N = 55; Point P[N]; double R; int n; double CommonArea(Circle C) { double res = 0; for (int i = 0; i < n; i++) res += GetCommonArea(C.c, P[i], P[i + 1], C); return res; } int main() { int cas = 1; while (scanf(%d, &n) == 1 && n) { scanf(%lf, &R); for (int i = 0; i < n; i++) scanf(%lf%lf, &P[i].x, &P[i].y); P[n] = P[0]; double l = 0, r = 1e9; while (dcmp(r - l) > 0) { double mid = (l + r) * 0.5; if (dcmp(CommonArea(Circle(Point(0, 0), mid)) - R) >= 0) r = mid; else l = mid; } printf(Case %d: %.2lf , cas++, l); } return 0; }
       
      
     
    
   

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