UVa - 12617 - How Lader

先上题目：

How Lader

Lader is a game that is played in a regular hexagonal board (all sides equal, all angles are also equal). The game is much similar as pool game. But there is only one hole that is situated in the center of the hexagon. The position of the board is given by a 2D co-ordinate system. The top and bottom sides of the hexagon are parallel to x axis. The center of the hexagonal board is situated at (0,0).

$\textstyle \parbox{.5\textwidth}{ \begin{center} \mbox{} \epsfbox{p12617a.eps} \end{center}}$ $\textstyle \parbox{.49\textwidth}{ \begin{center} \mbox{} \epsfbox{p12617b.eps} \end{center}}$

You are trying to hit the ball B1 and the direction of hitting is from B1 to B2. After you have hit the ball B1, it starts reflecting on the walls of the hexagonal Lader board. The initial speed of the ball is given. When a ball hits a wall, its speed decreases by 1 unit/second. The ball stops when its' speed becomes $\le$ 0unit/second.

You have to determine the final speed of the ball when it falls through the hole. If the ball stops before reaching the hole, print `Stops'. In this problem assume the followings:

There is no loss of speed while rolling freely on the board.
The radius of the ball is so small that you can consider it as a point.
You may consider the ball fallen in the hole, if at any point the ball is situated at a distance closer than r + 10^-6 units from the center of the hole, where r is the radius of the hole.
The reflection happens according to the standard reflection rule (incident angle = reflection angle, with respect to the side of the hexagon) except for the case when it hits the corner. That case is described in 5-th rule.
If a ball reaches at the corner (intersection point of two sides), its speed decreases by 2 (it is assumed that it hits both the walls) and it comes back along the line it hits that corner. If a ball with speed 1 hits the corner, it stops there.

The picture on the right above shows the movements of a ball on a Lader board. The numbers written denote the order of appearance.

Input

The first line of the input denotes T ( 1 $\le$ T $\le$ 150), the number of test cases to follow. Each test case consists of a 6 integers, s ( 0 < s < 150), x1, y1, x2, y2, r, t (1 $\le$ t $\le$ 500). Here, s denotes the length of sides of the hexagon centered at (0,0).

(x1, y1) and (x2, y2) denote the position of ball B1 and ball B2 respectively. The balls will be strictly inside the hexagonal board. r denotes the radius of the hole, centered at (0,0). The hole resides strictly inside the hexagonal board. t denotes the initial speed of the ball.

Output

For each input, you have to print the case number first, followed by the terminal speed when it falls in the hole. If the ball stops before falling in the hole, print `Stops'.

Sample Input

4
80 10 0 20 0 5 200
51 7 4 0 9 5 1
55 -5 8 -6 7 8 104
12 1 0 0 -1 1 271

Sample Output

Case 1: 198
Case 2: Stops
Case 3: 99

Problemsetter: Anna Fariha
Special Thanks: Md. Mahbubul Hasan

　　题意：给你一个正六边形，中间有一个半径为R的洞，现在有一个球b1给他一个方向向量以及速度。球每一次碰撞六边形的边速度会减1，如果撞到角的话速度会减2，问你当球掉进洞里的时候速度是多少，如果还没有掉进洞里速度就小于等于0的话就输出"Stops"。

　　几何+模拟。

　　判断射线是否穿过点，射线与线段相交，射线与圆的交点，以及向量的反射。这要这些都解决的话就没有太多问题了。

　　关于射线穿过点，射线与线段相交等，可以看一下该博客的一份几何模板。这里讲一下射线与圆的相交判断，射线与圆相交或者相切，可以用过解二元一次方程得到，根据判别式的值我们可以判断蛇蝎和圆的相交情况。这与向量的反射这里给出一条公式:v'=v+N*2*fabs(Dot(v,N)),其中这里v是入射向量v'是出射向量，N是反射面的法线向量，Dot(v,N)是点积。这里需要注意的是求点积以后需要求绝对值，因为这里求点积的作用是为了求向量在法线上的投影长度，所以需要转成正数。

　　需要注意的地方是对于起点来说，如果一开始它就在原的里面或者边上的话，那它就一开始就可以输出结果了(特别注意的是在边上的情况)。

上代码：

  1 #include <cstdio>
  2 #include <cstring>
  3 #include <cmath>
  4 #include <algorithm>
  5 #define MAX 10
  6 using namespace std;
  7 
  8 const double PI=2*acos(0);
  9 const double der60=PI/3;
 10 const double eps=1e-6;
 11 const double sqrt3=sqrt(3);
 12 
 13 int dcmp(double x){
 14     if(fabs(x)<eps) return 0;
 15     return x>0 ? 1 : -1;
 16 }
 17 
 18 typedef struct Point{
 19     double x,y;
 20     Point(double x=0,double y=0):x(x),y(y){}
 21 }Point;
 22 typedef Point Vector;
 23 Vector operator + (Point A,Point B){ return Vector(A.x+B.x,A.y+B.y);}
 24 Vector operator - (Point A,Point B){ return Vector(A.x-B.x,A.y-B.y);}
 25 Vector operator * (Point A,double e){ return Vector(A.x*e,A.y*e);}
 26 Vector operator / (Point A,double e){ return Vector(A.x/e,A.y/e);}
 27 bool operator ==  (Point A,Point B){ return dcmp(A.x-B.x)==0 && dcmp(A.y-B.y)==0;}
 28 double Dot(Vector A,Vector B){ return A.x*B.x+A.y*B.y;}
 29 double Cross(Vector A,Vector B){ return A.x*B.y-A.y*B.x;}
 30 double Length(Vector A){ return sqrt(Dot(A,A));}
 31 
 32 Vector Rotate(Vector A,double rad){
 33     return Vector(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad));
 34 }
 35 Vector Normal(Vector A){
 36     double L=Length(A);
 37     if(dcmp(L)==0) return Vector(0,0);
 38     return Vector(-A.y/L,A.x/L);
 39 }
 40 Point p[6],b1,b2,st;
 41 Vector di;
 42 double s;
 43 int ti;
 44 
 45 typedef struct Circle{
 46     Point c;
 47     double r;
 48 }Circle;
 49 Circle cen;
 50 
 51 int getLCI(Point p0,Vector v,double &t1,double &t2){
 52     double a=v.x;   double b=p0.x-cen.c.x;
 53     double c=v.y;   double d=p0.y-cen.c.y;
 54     double e=a*a+c*c;   double f=2*(a*b+c*d);   double g=b*b+d*d-cen.r*cen.r;
 55     double delta=f*f-4*e*g;
 56     if(dcmp(delta)<0) return 0;
 57     if(dcmp(delta)==0){
 58         t1=t2=-f/(2*e);
 59         return 1;
 60     }
 61     t1=(-f-sqrt(delta))/(2*e);
 62     t2=(-f+sqrt(delta))/(2*e);
 63     return 2;
 64 }
 65 bool OnSegment(Point p0,Point a1,Point a2){
 66     return (dcmp(Cross(a1-p0,a2-p0))==0 && dcmp(Dot(a1-p0,a2-p0))<0);
 67 }
 68 
 69 bool isPar(Point a1,Point a2){
 70     Vector v=a2-a1;
 71     v=v/Length(v);
 72     if(v==di || (v*-1)==di) return 1;
 73     return 0;
 74 }
 75 
 76 
 77 Point GLI(Point P,Vector v,Point Q,Vector w){
 78     Vector u=P-Q;
 79     double t=Cross(w,u)/Cross(v,w);
 80     return P+v*t;
 81 }
 82 
 83 bool isOnLine(Point e){
 84     Vector u=e-st;
 85     u=u/Length(u);
 86     if(u==di) return 1;
 87     return 0;
 88 }
 89 
 90 int solve(){
 91     int ans=ti;
 92     double t1,t2;
 93     Point tt;
 94     Vector sv,ndi,normal;
 95     bool f;
 96     while(ans>0){
 97         if(getLCI(st,di,t1,t2)>0){
 98             if(t1>=0 || t2>=0) return ans;
 99         }
100         f=0;
101         for(int i=0;i<6;i++){
102             if(isOnLine(p[i])){
103                 st=p[i]; di=di*-1;
104                 ans-=2; f=1;
105                 break;
106             }
107         }
108         if(f) continue;
109         for(int i=0;i<6;i++){
110             if(OnSegment(st,p[i],p[(i+1)%6])) continue;
111             if(isPar(p[i],p[(i+1)%6])) continue;
112             sv=p[(i+1)%6]-p[i];
113             tt=GLI(st,di,p[i],sv);
114             if(isOnLine(tt) && OnSegment(tt,p[i],p[(i+1)%6])){
115                 st=tt;
116                 normal=Normal(sv);
117                 ndi=di+normal*2*fabs(Dot(di,normal));
118                 di=ndi;
119                 di=di/Length(di);
120                 ans--;
121                 break;
122             }
123         }
124     }
125     return 0;
126 }
127 
128 
129 int main()
130 {
131     int t,ans;
132     Vector e;
133     //freopen("data.txt","r",stdin);
134     scanf("%d",&t);
135     for(int z=1;z<=t;z++){
136         scanf("%lf %lf %lf %lf %lf %lf %d",&s,&b1.x,&b1.y,&b2.x,&b2.y,&cen.r,&ti);
137         di=b2-b1;
138         di=di/Length(di);
139         st=b1;
140         cen.c.x=cen.c.y=0;
141         p[0].x=-s;   p[0].y=0;
142         p[1].x=-s/2; p[1].y=-s*sqrt3/2;
143         p[2].x=s/2;  p[2].y=-s*sqrt3/2;
144         p[3].x=s;    p[3].y=0;
145         p[4].x=s/2;  p[4].y=s*sqrt3/2;
146         p[5].x=-s/2; p[5].y=s*sqrt3/2;
147 //        for(int i=1;i<6;i++){
148 //            p[i]=Rotate(p[i-1],der60);
149 //        }
150         ans=solve();
151         printf("Case %d: ",z);
152         if(ans) printf("%d\n",ans);
153         else printf("Stops\n");
154     }
155     return 0;
156 }

/*12617*/

posted @ 2014-08-30 19:11 海拉鲁的林克阅读(301) 评论(0) 编辑收藏举报

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