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 }
