csu1377Putter && HOJ12816

链接：（csu）http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1377

　　　（HOJ）http://49.123.82.55/online/?action=problem&type=list&courseid=0&querytext=&pageno=57

题意：

　　给定凸多边形的点（按顺时针），多边形内一点沿某个方向运动，碰到边进行镜面反射，问在一条边不碰超过两次的情况下，有多少种不同的碰撞方法。多边形最多8条边。

思路：

　　枚举每一种情况，合理就对结果+1，枚举用dfs轻松加愉快。本题关键在于求出镜像点，然后以镜像点出发，对边进行考察，若能直线到达，则该点能被反射到，否则不能。

Code：

 #include <iostream>

 #include <cstdio>

 #include <cmath>

 #include <cstring>

 #include <algorithm>

 #define op operator

 #define cp const P&

 #define cn const

 #define db double

 #define rt return

 using namespace std;

 cn db eps = 1e-;

 cn db pi = acos(-1.0);

 inline int sig(db x) {return (x>eps) - (x<-eps);}

 struct P{

     db x, y;

     P(db a = 0.0, db b = 0.0) : x(a), y(b) {}

     void in() { scanf("%lf %lf", &x, &y);}

     void out(){ printf("%lf %lf\n", x, y);}

     bool op<(cp a)cn {return sig(x-a.x) ? sig(x-a.x) : sig(y-a.y);}

     P op-(cp a)cn {return P(x-a.x, y-a.y);}

     P op+(cp a)cn {return P(x+a.x, y+a.y);}

     db op^(cp a)cn {return x*a.y - y*a.x;}

     db cross(P a, P b) {return (a-*this) ^ (b-*this);}

     db op*(cp a)cn {return x*a.x + y*a.y;} //点积

     db dot(P a, P b) {return (a-*this) * (b-*this);}

     P op*(cn db &a)cn {return P(a*x, a*y);}

     bool op/(cp a)cn {return sig(x*a.y - y*a.x) == ;}

     db dis() {return sqrt(x*x + y*y);}

     db dis(P a) {return (a-*this).dis();}

     P T() {return P(-y, x);}

     P roate(db d) {

         return P(x*cos(d) - y*sin(d), y*cos(d) + x*sin(d));

     }

     bool on(P a, P b) {return !sig(cross(a, b)) && sig(dot(a, b)) <= ;} //点在线段上

     P Mirror_P(P a, P b) {

         P v1 = *this - a, v2 = b - a;

         db w = acos(v1*v2 / v1.dis() / v2.dis());

         int f = sig(v1^v2);

         w = *w*f;

         rt a + v1.roate(w);

     }

 }p[], q;

 inline P inset(P a1, P b1, P a2, P b2) {

     db u = (b1^a1), z = b2^a2, w = (b1-a1) ^ (b2-a2);

     P v;

     v.x = ((b2.x-a2.x)*u - (b1.x-a1.x)*z) / w;

     v.y = ((b2.y-a2.y)*u - (b1.y-a1.y)*z) / w;

     rt v;

 }

 bool vis[];

 int ans, n;

 bool if_insight(P q, int i, P a, P b) {

     if(sig(q.cross(b, p[i])) <=  && sig(q.cross(b, p[i+])) <= ) rt false;

     if(sig(q.cross(a, p[i])) >=  && sig(q.cross(a, p[i+])) >= ) rt false;

     rt true;

 }

 void dfs(P q, int k, int m, P a, P b) {

     if(vis[k]) rt ;

     if(m == n) { ++ans; rt ; }

     vis[k] = ;

     q = q.Mirror_P(p[k], p[k+]);

     for(int i = ; i < n; ++i) {

         if(i != k && if_insight(q, i, a, b)) {

             P c , d;

             if(sig(q.cross(b, p[i])) < ) c = inset(b, q, p[i], p[i+]);

             else c = p[i];

             if(sig(q.cross(a, p[i+])) > ) d = inset(a, q, p[i], p[i+]);

             else d = p[i+];

             dfs(q, i, m+, c, d);

         }

     }

     vis[k] = ;

 }

 void solve() {

     p[n] = p[];

     ans = ;

     for(int i = ; i < n; ++i)

         dfs(q, i, , p[i], p[i+]);

     printf("%d\n", ans);

     rt ;

 }

 int main()

 {

     while(scanf("%d", &n), n) {

         q.in();

         for(int i = ; i < n; ++i) p[i].in();

         memset(vis, , sizeof vis);

         solve();

     }

     return ;

 }