NWERC 2015 Problem H: Hole in One

暴力计算几何。。

#include <bits/stdc++.h>
using namespace std;
const double eps = 1e-10;
inline int sgn(double x)
{
    if (fabs(x) < eps)
        return 0;
    if (x < 0)
        return -1;
    return 1;
}
struct Point
{
    double x, y;
    Point() {}
    Point(double x, double y): x(x), y(y) {}
    void input()
    {
        scanf("%lf%lf", &x, &y);
    }
    double operator^(const Point &b)const
    {
        return x * b.y - y * b.x;
    }
    double operator*(const Point &b)const
    {
        return x * b.x + y * b.y;
    }
    Point operator-(const Point &b)const
    {
        return Point(x - b.x, y - b.y);
    }
} hole, a[15], b[15];
struct Line
{
    Point s, e;
    Line() {}
    Line(Point s, Point e): s(s), e(e) {}
    int segcrossseg(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);
    }
} wall[15];
int n;
vector<int> per;
int vis[15];
inline bool gao()
{
    memset(vis, 0, sizeof(vis));
    double xx = 0, yy = 0;
    double nowx = 0, nowy = 0;
    int flagx = 0, flagy = 0;
    int bx = 0, by = 0;
    /*判断反射方向，求反射长度*/
    for (int j = 0; j < (int)per.size(); ++j)
    {
        int id = per[j];
        /*讨论竖直水平*/
        if (sgn(wall[id].s.x - wall[id].e.x) == 0)
        {
            if (flagx == 0)
            {
                flagx = sgn(wall[id].s.x - nowx);
                if (sgn(wall[id].s.x - nowx) == 0 ) return 0;
                bx = flagx;
            }
            else
            {
                if (flagx == sgn(wall[id].s.x - nowx)) return 0;
                if (sgn(wall[id].s.x - nowx) == 0 ) return 0;
                flagx = -flagx;
            }
            xx += fabs(wall[id].s.x - nowx);
            nowx = wall[id].s.x;
        }
        else if (sgn(wall[id].s.y - wall[id].e.y) == 0)
        {
            if (flagy == 0)
            {
                flagy = sgn(wall[id].s.y - nowy);
                if (sgn(wall[id].s.y - nowy) == 0 ) return 0;
                by = flagy;
            }
            else
            {
                if (flagy == sgn(wall[id].s.y - nowy)) return 0;
                if (sgn(wall[id].s.y - nowy) == 0 ) return 0;
                flagy = -flagy;
            }
            yy += fabs(wall[id].s.y - nowy);
            nowy = wall[id].s.y;
        }
    }
    /*判断最后一下方向*/
    if (bx == 0) bx = 1;
    if (by == 0) by = 1;
    // 默认右上角
    xx += fabs(hole.x - nowx);
    yy += fabs(hole.y - nowy);
    if (sgn(hole.x - nowx) == flagx && flagx != 0) return 0;
    if (sgn(hole.y - nowy) == flagy && flagy != 0) return 0;
    //  回不到洞
    /*遍历反射过程*/
    double tmp = yy / xx;
    int tmpflag = 0;
    if (sgn(xx) == 0) tmpflag = 1;
    nowx = 0, nowy = 0;
    for (int j = 0; j < (int)per.size(); ++j)
    {
        int id = per[j];
        if (sgn(wall[id].s.x - wall[id].e.x) == 0)
        {
            double tmpx = fabs(wall[id].s.x - nowx);
            double zz;
            if (tmpflag == 0)zz = tmpx * tmp * by;
            else zz = 0;
            Point pp(wall[id].s.x, nowy + zz);
            Line ll(pp, Point(nowx, nowy));
            if (ll.segcrossseg(wall[id]))
            {
                vis[id] = 1;
                for (int i = 0 ; i < n; i++)
                {
                    if (vis[i]) continue;
                    if(ll.segcrossseg(wall[i])) return 0;
                }
            }
            else return 0;
            bx = -bx;
            nowx = pp.x ;
            nowy = pp.y;
        }
        else
        {
            double tmpy = fabs(wall[id].s.y - nowy);
            double zz;
            if (tmpflag == 0)zz = tmpy / tmp * bx;
            else zz = 0;
            Point pp(nowx + zz, wall[id].s.y);
            Line ll(pp, Point(nowx, nowy));
            if (ll.segcrossseg(wall[id]))
            {
                vis[id] = 1;
                for (int i = 0 ; i < n; i++)
                {
                    if (vis[i]) continue;
                    if(ll.segcrossseg(wall[i])) return 0;
                }
            }
            else return 0;
            by = -by;
            nowx = pp.x ;
            nowy = pp.y;
        }
    }
    /*最后一次相交*/
    double tmpx = fabs(hole.x - nowx);
    double tmpy = fabs(hole.y - nowy);
    double zz;
    if (tmpflag == 0)zz = tmpx * tmp * by;
    else zz = 0;
    Point pp(hole.x, nowy + zz);
    if (sgn(pp.y - (nowy + zz)) != 0) return 0;
    Line ll(pp, Point(nowx, nowy));
    for (int i = 0 ; i < n; i++)
    {
        if (vis[i]) continue;
        if(ll.segcrossseg(wall[i])) return 0;
    }
    return 1;
}

inline int solve(int mask)
{
    per.clear();
    for (int i = 0; i < n; ++i)
        if ((mask >> i) & 1)
            per.push_back(i);
    int ret = 0;
    do
    {
        if (gao()) return per.size();
    }
    while (next_permutation(per.begin(), per.end()));
    return -1;
}
int main()
{
    scanf("%d", &n);
    hole.input();
    for (int i = 0; i < n; ++i)
    {
        a[i].input();
        b[i].input();
    }
    if (sgn(hole.x) < 0)
    {
        for (int i = 0; i < n; ++i)
        {
            a[i].x = -a[i].x;
            b[i].x = -b[i].x;
        }
        hole.x = -hole.x;
    }
    if (sgn(hole.y) < 0)
    {
        for (int i = 0; i < n; ++i)
        {
            a[i].y = -a[i].y;
            b[i].y = -b[i].y;
        }
        hole.y = -hole.y;
    }
    for (int i = 0; i < n; ++i)
        wall[i] = Line(a[i], b[i]);
    int st = 1 << n, res = -1;
    for (int i = 0; i < st; ++i)
        res = max(res, solve(i));
    if (res == -1) puts("impossible");
    else printf("%d
", res);
    return 0;
}