简单几何(直线与圆的交点) ZOJ Collision 3728-白红宇

简单几何(直线与圆的交点) ZOJ Collision 3728
阅读量：5025 次
发布时间：2019-06-12
本文共 6162 字，大约阅读时间需要 20 分钟。
题意：有两个一大一小的同心圆，圆心在原点，大圆外有一小圆，其圆心有一个速度(vx, vy)，如果碰到了小圆会反弹，问该圆在大圆内运动的时间
分析：将圆外的小圆看成一个点，判断该直线与同心圆的交点，根据交点个数计算时间。用到了直线的定义，圆的定义，直线与圆交点的个数。
/************************************************* Author        :Running_Time* Created Time  :2015/10/24 星期六 16:14:33* File Name     :C.cpp ************************************************/#include 
     
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                      using namespace std;#define lson l, mid, rt << 1#define rson mid + 1, r, rt << 1 | 1typedef long long ll;const int N = 1e5 + 10;const int INF = 0x3f3f3f3f;const int MOD = 1e9 + 7;const double EPS = 1e-10;int dcmp(double x) { //三态函数，减少精度问题 if (fabs (x) < EPS) return 0; else return x < 0 ? -1 : 1;}struct Point { //点的定义 double x, y; Point (double x=0, double y=0) : x (x), y (y) {} Point operator + (const Point &r) const { //向量加法 return Point (x + r.x, y + r.y); } Point operator - (const Point &r) const { //向量减法 return Point (x - r.x, y - r.y); } Point operator * (double p) { //向量乘以标量 return Point (x * p, y * p); } Point operator / (double p) { //向量除以标量 return Point (x / p, y / p); } bool operator < (const Point &r) const { //点的坐标排序 return x < r.x || (x == r.x && y < r.y); } bool operator == (const Point &r) const { //判断同一个点 return dcmp (x - r.x) == 0 && dcmp (y - r.y) == 0; }};typedef Point Vector; //向量的定义Point read_point(void) { //点的读入 double x, y; scanf ("%lf%lf", &x, &y); return Point (x, y);}double polar_angle(Vector A) { //向量极角 return atan2 (A.y, A.x);}double dot(Vector A, Vector B) { //向量点积 return A.x * B.x + A.y * B.y;}double cross(Vector A, Vector B) { //向量叉积 return A.x * B.y - A.y * B.x;}double length(Vector A) { //向量长度，点积 return sqrt (dot (A, A));}double angle(Vector A, Vector B) { //向量转角，逆时针，点积 return acos (dot (A, B) / length (A) / length (B));}double area_triangle(Point a, Point b, Point c) { //三角形面积，叉积 return fabs (cross (b - a, c - a)) / 2.0;}Vector rotate(Vector A, double rad) { //向量旋转，逆时针 return Vector (A.x * cos (rad) - A.y * sin (rad), A.x * sin (rad) + A.y * cos (rad));}Vector nomal(Vector A) { //向量的单位法向量 double len = length (A); return Vector (-A.y / len, A.x / len);}Point point_inter(Point p, Vector V, Point q, Vector W) { //两直线交点，参数方程 Vector U = p - q; double t = cross (W, U) / cross (V, W); return p + V * t;}double dis_to_line(Point p, Point a, Point b) { //点到直线的距离，两点式 Vector V1 = b - a, V2 = p - a; return fabs (cross (V1, V2)) / length (V1);}double dis_to_seg(Point p, Point a, Point b) { //点到线段的距离，两点式 if (a == b) return length (p - a); Vector V1 = b - a, V2 = p - a, V3 = p - b; if (dcmp (dot (V1, V2)) < 0) return length (V2); else if (dcmp (dot (V1, V3)) > 0) return length (V3); else return fabs (cross (V1, V2)) / length (V1);}Point point_proj(Point p, Point a, Point b) { //点在直线上的投影，两点式 Vector V = b - a; return a + V * (dot (V, p - a) / dot (V, V));}bool inter(Point a1, Point a2, Point b1, Point b2) { //判断线段相交，两点式 double c1 = cross (a2 - a1, b1 - a1), c2 = cross (a2 - a1, b2 - a1), c3 = cross (b2 - b1, a1 - b1), c4 = cross (b2 - b1, a2 - b1); return dcmp (c1) * dcmp (c2) < 0 && dcmp (c3) * dcmp (c4) < 0;}bool on_seg(Point p, Point a1, Point a2) { //判断点在线段上，两点式 return dcmp (cross (a1 - p, a2 - p)) == 0 && dcmp (dot (a1 - p, a2 - p)) < 0;}double area_poly(Point *p, int n) { //多边形面积 double ret = 0; for (int i=1; i
                      
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                         &P) { sort (P.begin (), P.end ()); P.erase (unique (P.begin (), P.end ()), P.end ()); //预处理，删除重复点 int n = P.size (), m = 0; vector
                        
                          ret (n + 1); for (int i=0; i
                         
                           1 && cross (ret[m-1]-ret[m-2], P[i]-ret[m-2]) < 0) m--; ret[m++] = P[i]; } int k = m; for (int i=n-2; i>=0; --i) { while (m > k && cross (ret[m-1]-ret[m-2], P[i]-ret[m-2]) < 0) m--; ret[m++] = P[i]; } if (n > 1) m--; ret.resize (m); return ret;} 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 r; Line () {} Line (const Point &p, const Vector &v) : p (p), v (v) { r = polar_angle (v); } Point point(double a) { return p + v * a; }};/* 直线相交求交点，返回交点个数，交点保存在P中*/int line_cir_inter(Line L, Circle C, double &t1, double &t2, vector
                          
                            &P) { 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); P.push_back (L.point (t1)); return -1; } t1 = (-f - sqrt (delta)) / (2 * e); P.push_back (L.point (t1)); t2 = (-f + sqrt (delta)) / (2 * e); P.push_back (L.point (t2)); if (dcmp (t1) < 0 || dcmp (t2) < 0) return 0; return 2;} /* 两圆相交求交点，返回交点个数。交点保存在P中*/int cir_cir_inter(Circle C1, Circle C2, vector
                           
                             &P) { double d = length (C1.c - C2.c); if (dcmp (d) == 0) { if (dcmp (C1.r - C2.r) == 0) return -1; //两圆重叠 else return 0; } if (dcmp (C1.r + C2.r - d) < 0) return 0; if (dcmp (fabs (C1.r - C2.r) - d) < 0) return 0; double a = polar_angle (C2.c - C1.c); double da = acos ((C1.r * C1.r + d * d - C2.r * C2.r) / (2 * C1.r * d)); //C1C2到C1P1的角？ Point p1 = C1.point (a - da), p2 = C2.point (a + da); P.push_back (p1); if (p1 == p2) return 1; else P.push_back (p2); return 2;}int main(void) { Circle C1, C2; Point a, b; double Rm, R, r, x, y, vx, vy; while (scanf ("%lf%lf%lf%lf%lf%lf%lf", &Rm, &R, &r, &x, &y, &vx, &vy) == 7) { Rm += r; R += r; C1 = Circle (Point (0, 0), Rm); C2 = Circle (Point (0, 0), R); a = Point (x, y); b = Point (vx, vy); vector
                            
                              P1, P2; P1.clear (); P2.clear (); double t1, t2, t3, t4, ans, speed = sqrt (vx * vx + vy * vy); int num1 = line_cir_inter (Line (a, b), C1, t1, t2, P1); int num2 = line_cir_inter (Line (a, b), C2, t3, t4, P2); if (num2 == 2) { if (num1 == 2) { double len = length (P2[0] - P2[1]) - length (P1[0] - P1[1]); ans = len / speed; } else { ans = length (P2[0] - P2[1]) / speed; } } else { ans = 0.0; } printf ("%.4f\n", ans); } return 0;}
转载于:https://www.cnblogs.com/Running-Time/p/4907417.html
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