2013 ACM区域赛长沙 C Collision HDU 4793
题意：在平面上0,0点，有一个半径为R的圆形区域，并且在0,0点固定着一个半径为RM（<R）的圆形障碍物，现在圆形区域外x,y，有一个半径 为r的，并且速度为vx,vy的硬币，如果硬币碰到了障碍物，将会保持原有的速度向反射的方向继续前进，现在给出R,RM,r,x,y,vx,vy，问硬币的任意部分在圆形区域中滑行多少时间？
思路：首先把R,RM加上r，就可以把硬币看做一个点来讨论了，然后算一下射线与这两个圆交点的个数，记作C1，C2，CM1，CM2,若与两个圆的交点数都是2，答案就是dis(c1,c2)-dis(cm1,cm2)之后再除以合成后的速度，若与大圆的交点数为2，小圆的交点数是0或者1，答案就是 dis（c1,c2）/速度，否则肯定是0了.
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<string> 5 #include<algorithm> 6 #include<map> 7 #include<queue> 8 #include<set> 9 #include<stack> 10 #include<cmath> 11 #include<vector> 12 #define inf 0x3f3f3f3f 13 #define Inf 0x3FFFFFFFFFFFFFFFLL 14 #define eps 1e-9 15 #define pi acos(-1.0) 16 using namespace std; 17 typedef long long ll; 18 19 int dcmp(double x) { 20 return (x>eps)-(x<-eps); 21 } 22 23 typedef struct Point { 24 double x, y; 25 Point(double x = 0, double y = 0) : x(x), y(y) {} 26 Point operator+(const Point& p) const { 27 return Point(x+p.x, y+p.y ); 28 } 29 Point operator-(const Point& p) const { 30 return Point(x-p.x, y-p.y ); 31 } 32 Point operator*(const double d) const { 33 return Point(x*d, y*d ); 34 } 35 Point operator/(const double d) const { 36 return Point(x/d, y/d ); 37 } 38 void read() { 39 scanf("%lf%lf", &x, &y); 40 } 41 } Vector; 42 43 inline double Dis(const Point& a, const Point& b) { 44 return sqrt((a.x-b.x)*(a.x-b.x) + (a.y-b.y)*(a.y-b.y)); 45 } 46 47 struct Line { 48 Point P; 49 Vector v; 50 double ang; 51 Line() {} 52 Line(const Point& P, const Vector& v):P(P),v(v) { 53 ang = atan2(v.y,v.x); 54 } 55 bool operator<(const Line& L) const { 56 return ang<L.ang; 57 } 58 Point point(double t) { 59 return P+v*t; 60 } 61 }; 62 63 struct Circle { 64 Point c; 65 double r; 66 Circle() {} 67 }; 68 69 int GetLineCircleIntersection(Line L, Circle C, double &t1, double &t2, vector<Point> &sol) { 70 double a = L.v.x, b = L.P.x-C.c.x, c = L.v.y, d = L.P.y - C.c.y; 71 double e = a*a+c*c, f = 2*(a*b+c*d), g = b*b+d*d-C.r*C.r; 72 double delta = f*f-4*e*g; 73 if(dcmp(delta)<0) return 0; 74 if(dcmp(delta)==0) { 75 t1 = t2 = -f/(2*e); 76 sol.push_back(L.point(t1)); 77 return 1; 78 } 79 t1 = (-f-sqrt(delta))/(2*e); 80 sol.push_back(L.point(t1)); 81 t2 = (-f+sqrt(delta))/(2*e); 82 sol.push_back(L.point(t2)); 83 if(dcmp(t1)<0 || dcmp(t2)<0) return 0; 84 return 2; 85 } 86 87 double R,RM,r,x,y,vx,vy; 88 int n,m,k; 89 90 int main() { 91 // freopen("in.txt","r",stdin); 92 while(~scanf("%lf%lf%lf%lf%lf%lf%lf",&RM,&R,&r,&x,&y,&vx,&vy)) { 93 Line l1(Point(x,y),Point(vx,vy)); 94 Circle c1,c2; 95 c1.r=RM+r; 96 c1.c.x=c1.c.y=0.0; 97 c2.c.x=c2.c.y=0.0; 98 c2.r=R+r; 99 double db1=0.0,db2=0.0; 100 vector<Point> crs1; 101 vector<Point> crs2; 102 int num2=GetLineCircleIntersection(l1,c1,db1,db2,crs2); 103 int num1=GetLineCircleIntersection(l1,c2,db1,db2,crs1); 104 // printf("num1:%d num2:%d\n",num1,num2); 105 double len=0.0,ans=0.0; 106 double vv=sqrt(vx*vx+vy*vy); 107 if (num1==2 && num2==2) { 108 len=Dis(crs1[0],crs1[1])-Dis(crs2[0],crs2[1]); 109 printf("%.4lf\n",len/vv); 110 } else if (num1==2) { 111 len=Dis(crs1[0],crs1[1]); 112 printf("%.4lf\n",len/vv); 113 } else { 114 printf("0.0000\n"); 115 } 116 } 117 return 0; 118 }
