hdu 2872 Another Snake 爆搜 判断射线与线段相交

题意： 给出n各点，有一条蛇从原点开始走。每次只能向左 [0,180) 转。并且不能和原路径相交。走到最后一个点后会一直往前走。
问最多能走到的点数


假设现在走到的点为p，下一个点q。如果射线pq 与之前的路径相交了，由于只能左转，之后无论怎么走，最终必然会与路径相交，因此此时走pq是不合法的。
合理的方案数是很少的，可以暴力的搜索。

每次只要判断 射线是否与原路径相交，是否为向左转即可。

#include<stdio.h>#include<string.h>#include<ctype.h>#include<math.h>#include<string>#include<map>#include<vector>#include<queue>#include<algorithm>using namespace std;void fre(){freopen("t.txt","r",stdin);}#define ls o<<1#define rs o<<1|1typedef long long LL;typedef unsigned long long UL;typedef unsigned int UI;const int MAXN = 1<<30;const int N = 50000;const double eps = 1e-16;int sgn(double x){    if(fabs(x) < eps) return 0;    if(x < 0) return -1;    else return 1;}struct Point{    double x,y;    Point(){}    Point(double _x, double _y)    {        x = _x; y = _y;    }    void input()    {        scanf("%lf%lf",&x,&y);    }    Point operator -(const Point &b)const    {        return Point(x-b.x,y-b.y);    }    double operator *(const Point &b)const    {        return x*b.x + y*b.y;    }    double operator ^(const Point &b)const    {        return x*b.y - y*b.x;    }}p[20],path[20];struct Line{    Point s,e;    Line(){}    Line(Point _s,Point _e)    {        s = _s; e = _e;    }    int segXseg(Line v)    {        int d1 = sgn((e-s)^(v.s-s));        int d2 = sgn((e-s)^(v.e-s));        int d3 = sgn((v.e-v.s)^(s-v.s));        int d4 = sgn((v.e-v.s)^(e-v.s));        if( (d1^d2)==-2 && (d3^d4)==-2 ) return 2;        return (d1 == 0 && sgn((v.s-s)*(v.s-e)) <= 0) ||            (d2 == 0 && sgn((v.e-s)*(v.e-e)) <= 0) ||            (d3 == 0 && sgn((s-v.s)*(s-v.e)) <= 0) ||            (d4 == 0 && sgn((e-v.s)*(e-v.e)) <= 0);    }};int n,ans,t;bool vis[20];bool ok(int u){    double d1 = (p[u]-path[t]) ^ (path[t] - path[t-1]), d2 =(p[u]-path[t]) * (path[t] - path[t-1]);    if( !(d1 < -eps || (fabs(d1) < eps && d2 > eps)) ) return 0;    Line l = Line(path[t],p[u]);    l.e.x = l.s.x + (l.e.x - l.s.x) * 100000000.0;    l.e.y = l.s.y + (l.e.y - l.s.y) * 100000000.0;    for(int i = 1; i < t; ++i)    {        if(l.segXseg(Line(path[i],path[i-1]))) return 0;    }    return 1;}void dfs(){    for(int i = 1; i <= n; ++i)    {        if(ans == n) return;        if(!vis[i] && (t < 1 || ok(i)))        {            vis[i] = 1; path[++t] = p[i];            if(t > ans) ans = t;            dfs();            vis[i] = 0; t--;        }    }}int main(){    //fre();    while(~scanf("%d",&n))    {        memset(vis,0,sizeof(vis));        ans = t = 0;        for(int i = 1; i <= n; ++i) p[i].input();        path[0] = Point(0,0);        dfs();        printf("%d\n",ans);    }}