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:

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cmath>
 4 #include <cstring>
 5 #include <algorithm>
 6 #define op operator
 7 #define cp const P&
 8 #define cn const
 9 #define db double
10 #define rt return
11 using namespace std;
12 cn db eps = 1e-9;
13 cn db pi = acos(-1.0);
14 inline int sig(db x) {return (x>eps) - (x<-eps);}
15 
16 struct P{
17     db x, y;
18     P(db a = 0.0, db b = 0.0) : x(a), y(b) {}
19     void in() { scanf("%lf %lf", &x, &y);}
20     void out(){ printf("%lf %lf\n", x, y);}
21     bool op<(cp a)cn {return sig(x-a.x) ? sig(x-a.x) : sig(y-a.y);}
22     P op-(cp a)cn {return P(x-a.x, y-a.y);}
23     P op+(cp a)cn {return P(x+a.x, y+a.y);}
24     db op^(cp a)cn {return x*a.y - y*a.x;}
25     db cross(P a, P b) {return (a-*this) ^ (b-*this);}
26     db op*(cp a)cn {return x*a.x + y*a.y;} //点积
27     db dot(P a, P b) {return (a-*this) * (b-*this);}
28     P op*(cn db &a)cn {return P(a*x, a*y);}
29     bool op/(cp a)cn {return sig(x*a.y - y*a.x) == 0;}
30     db dis() {return sqrt(x*x + y*y);}
31     db dis(P a) {return (a-*this).dis();}
32     P T() {return P(-y, x);}
33     P roate(db d) {
34         return P(x*cos(d) - y*sin(d), y*cos(d) + x*sin(d));
35     }
36     bool on(P a, P b) {return !sig(cross(a, b)) && sig(dot(a, b)) <= 0;} //点在线段上
37     P Mirror_P(P a, P b) {
38         P v1 = *this - a, v2 = b - a;
39         db w = acos(v1*v2 / v1.dis() / v2.dis());
40         int f = sig(v1^v2);
41         w = 2*w*f;
42         rt a + v1.roate(w);
43     }
44 }p[10], q;
45 inline P inset(P a1, P b1, P a2, P b2) {
46     db u = (b1^a1), z = b2^a2, w = (b1-a1) ^ (b2-a2);
47     P v;
48     v.x = ((b2.x-a2.x)*u - (b1.x-a1.x)*z) / w;
49     v.y = ((b2.y-a2.y)*u - (b1.y-a1.y)*z) / w;
50     rt v;
51 }
52 bool vis[10];
53 int ans, n;
54 bool if_insight(P q, int i, P a, P b) {
55     if(sig(q.cross(b, p[i])) <= 0 && sig(q.cross(b, p[i+1])) <= 0) rt false;
56     if(sig(q.cross(a, p[i])) >= 0 && sig(q.cross(a, p[i+1])) >= 0) rt false;
57     rt true;
58 }
59 void dfs(P q, int k, int m, P a, P b) {
60     if(vis[k]) rt ;
61     if(m == n) { ++ans; rt ; }
62     vis[k] = 1;
63     q = q.Mirror_P(p[k], p[k+1]);
64     for(int i = 0; i < n; ++i) {
65         if(i != k && if_insight(q, i, a, b)) {
66             P c , d;
67             if(sig(q.cross(b, p[i])) < 0) c = inset(b, q, p[i], p[i+1]);
68             else c = p[i];
69             if(sig(q.cross(a, p[i+1])) > 0) d = inset(a, q, p[i], p[i+1]);
70             else d = p[i+1];
71 
72             dfs(q, i, m+1, c, d);
73         }
74     }
75     vis[k] = 0;
76 }
77 void solve() {
78     p[n] = p[0];
79     ans = 0;
80     for(int i = 0; i < n; ++i)
81         dfs(q, i, 1, p[i], p[i+1]);
82     printf("%d\n", ans);
83     rt ;
84 }
85 int main()
86 {
87     while(scanf("%d", &n), n) {
88         q.in();
89         for(int i = 0; i < n; ++i) p[i].in();
90         memset(vis, 0, sizeof vis);
91         solve();
92     }
93     return 0;
94 }
View Code
posted @ 2014-07-23 20:08  妮king狼  阅读(160)  评论(0编辑  收藏  举报