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csu1377Putter && HOJ12816

2024-07-13 16:41:15   216人阅读

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 double10 #define rt return11 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 }
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