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UVa - 12617 - How Lader

發布時間：2025/4/14 编程问答 39 豆豆

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先上題目：

??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).

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??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?(?1T150), 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?(1t500). 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

?　　題意：給你一個正六邊形，中間有一個半徑為R的洞，現在有一個球b1給他一個方向向量以及速度。球每一次碰撞六邊形的邊速度會減1，如果撞到角的話速度會減2，問你當球掉進洞里的時候速度是多少，如果還沒有掉進洞里速度就小于等于0的話就輸出"Stops"。

　　幾何+模擬。

　　判斷射線是否穿過點，射線與線段相交，射線與圓的交點，以及向量的反射。這要這些都解決的話就沒有太多問題了。

　　關于射線穿過點，射線與線段相交等，可以看一下該博客的一份幾何模板。這里講一下射線與圓的相交判斷，射線與圓相交或者相切，可以用過解二元一次方程得到，根據判別式的值我們可以判斷蛇蝎和圓的相交情況。這與向量的反射這里給出一條公式:v'=v+N*2*fabs(Dot(v,N)),其中這里v是入射向量v'是出射向量，N是反射面的法線向量，Dot(v,N)是點積。這里需要注意的是求點積以后需要求絕對值，因為這里求點積的作用是為了求向量在法線上的投影長度，所以需要轉成正數。

　　需要注意的地方是對于起點來說，如果一開始它就在原的里面或者邊上的話，那它就一開始就可以輸出結果了(特別注意的是在邊上的情況)。

上代碼：

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*/

轉載于:https://www.cnblogs.com/sineatos/p/3946749.html

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