2017 Multi-University Training Contest - Team 9 1004&&HDU 6164 Dying Light【数学+模拟】

Dying Light

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 513    Accepted Submission(s): 122

Problem Description

LsF
is visiting a local amusement park with his friends, and a mirror room
successfully attracts his attention. Inside the mirror room, there are n
plane mirrors standing vertically on the ground. They are placed
end-to-end and face-to-face so that if you overlook the room, you can
find a convex hull and the all the reflector surfaces are inside the
pattern. The height of the mirror is not important in this problem.
Due
to imperfect manufacturing techniques, mirrors can't reflect light
without lose of energy. Each mirror has a reflection efficiency k, which
means if the incident light's intensity is I, the reflected light's
intensity will be reduced to kI. The only exception could happen when
the light precisely goes to the two mirrors' junction. In that case, the
light will be completely absorbed instantly. Note the laws of
reflection of light applies in all other situations, that the angle of
incidence equals the angle of reflection.
Now LsF stands inside the
mirror hall, and shoots a laser beam paralleled to the ground using his
laser pointer. Unfortunately, his laser pointer can only shot laser
beams with intensity of 1. What's worse, a laser beam is considered
disappeared if its intensity is below 10−4. There's not much magnitude distance between the two numbers.
LsF wants to know how many touches can his laser beam make with mirrors before it disappears.

 

Input

The first line contains an integer n(3≤n≤1000), indicating the number of mirrors;
Then n lines follow. The ith line contains three real numbers xi,yi,ki(−109≤xi,yi≤109;0≤ki≤0.9), which means the ith mirror's one end is at position (xi,yi) and another end is at (xi+1mod n,yi+1mod n), and its reflectivity is ki.
Next there are two real numbers Vx,Vy(-109≤Vx,Vy≤109), indicating the initial direction vector of the laser beam.
LsF is standing at the origin (0, 0).

 

Output

Output an integer in one line, the number of touches the laser beam could make before it disappears.

 

Sample Input

4
1 2 0.5
-1 0 0.5
1 -2 0.5
3 0 0.5
0 1
4
1 1 0.5
-1 1 0.5
-1 -1 0.5
1 -1 0.5
1 1

Sample Output

14
1

Source

2017 Multi-University Training Contest - Team 9

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6164

分析：由于LsF一开始不再镜子上，按题面模拟即可。
由于反射率<=0.9 0.9^100<1e-4，所以反射次数不会超过100次。
每次暴力判断和哪个镜子相交，以及有没有在镜子焦点上。

下面给出AC代码：

 #include <cstring>
 #include <algorithm>
 #include <cstdio>
 #include <iostream>

 #define MAXN 5000
 #define eps 1e-9

 struct point
 {
     double x,y;
     point(double a = ,double b = )
     {
         x = a; y = b;
     }
     friend point operator + (point a,point b)
     {
         return point(a.x+b.x,a.y+b.y);
     }
     friend point operator - (point a,point b)
     {
         return point(a.x-b.x,a.y-b.y);
     }
     friend double operator ^ (point a,point b)
     {
         return a.x*b.y-a.y*b.x;
     }
     friend double operator * (point a,point b)
     {
         return a.x*b.x+a.y*b.y;
     }
     friend point operator * (point a,double b)
     {
         return point(a.x*b,a.y*b);
     }
     friend point operator * (double a,point b)
     {
         return point(a*b.x,a*b.y);
     }

 };

 struct line
 {
     point s,e;
     line(point a = point(,),point b = point(,))
     {
         s = a; e = b;
     }
 };

 point p[MAXN+];
 double c[MAXN+];
 int n;
 point s[];

 int sgn(double x)
 {
     if (x>eps) return ;
     if (x<-eps) return -;
     return ;
 }

 point Get_Intersect(line a,line b)
 {
     double u=(a.e-a.s)^(b.s-a.s);
     double v=(a.s-a.e)^(b.e-a.e);
     point p;
     p.x=(b.s.x*v+b.e.x*u)/(v+u);
     p.y=(b.s.y*v+b.e.y*u)/(v+u);
     return p;
 }

 int main()
 {
 //    freopen("input.txt","r",stdin);
     while(scanf("%d",&n)!=EOF)
     {for (int i=;i<n;i++) scanf("%lf%lf%lf",&p[i].x,&p[i].y,&c[i]);
     p[n] = p[];
     s[] = point(,);
     scanf("%lf%lf",&s[].x,&s[].y);

     double now = 1.0;
     int ans = ;
     bool flag = ;
     point temp;
     point temp2;
     line l1,l2,l3,l4;
     while (now > 1e-)
     {
         ans++;
         for (int i=;i<n;i++)
         {
             if (!sgn((p[i]-s[])^s[]))
             {
                 now = ;
                 flag = ;
                 break;
             }
         }
         if (!flag) break;
         for (int i=;i<n;i++)
         {
             if (sgn((p[i]-s[])^s[]) >  && sgn(s[]^(p[i+]-s[]))>)
             {
                 l1 = line(p[i+],p[i]);
                 l2 = line(s[],s[]+s[]);
                 temp = Get_Intersect(l1,l2);

                 l3 = line(temp,point(p[i+].y-p[i].y,p[i].x-p[i+].x)+temp);
                 l4 = line(s[],point(p[i].x-p[i+].x,p[i].y-p[i+].y)+s[]);

                 temp2 = Get_Intersect(l3,l4);
                 temp2 = *temp2-s[];
                 s[] = temp;
                 s[] = temp2-s[];
                 now *= c[i];
                 break;
             }
         }
     }
     printf("%d\n",ans);
     }
     return ;
 }