poj1066

题意：给出图中所有线段，和一个点，问这个点要出正方形至少穿越多少线。

分析：所有线段端点都在正方形周边线上。走弯路是无意义的，不会使穿过的墙减少，因为线段都是接边界的，无法绕过。我们枚举周边线上的出口，连直线，看最少有几个交点。因为只有跨越端点的时候才会改变交点数量。所以我们只需在每两个端点间枚举一个点。

#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
using namespace std;

#define maxn 50
#define zero(x) (((x)>0?(x):-(x))<eps)
#define eps 1.0E-8

struct Point
{
    double x, y;
    Point()
    {
    }
    Point(double xx, double yy) :
        x(xx), y(yy)
    {
    }
} point[maxn * 2], s;

struct Line
{
    Point a, b;
    Line()
    {
    }
    Line(Point aa, Point bb) :
        a(aa), b(bb)
    {
    }
} line[maxn];

int n, m, ans;

bool operator <(const Point &a, const Point &b)
{
    return atan2(a.y, a.x) < atan2(b.y, b.x);
}

double xmult(Point p1, Point p2, Point p0)
{
    return (p1.x - p0.x) * (p2.y - p0.y) - (p2.x - p0.x) * (p1.y - p0.y);
}

int same_side(Point p1, Point p2, Line l)
{
    return xmult(l.a, p1, l.b) * xmult(l.a, p2, l.b) > eps;
}

int dots_inline(Point p1, Line l)
{
    return zero(xmult(p1, l.a, l.b));
}

int dot_online_in(Point p, Line l)
{
    return zero(xmult(p, l.a, l.b)) && (l.a.x - p.x) * (l.b.x - p.x) < eps && (l.a.y - p.y) * (l.b.y - p.y) < eps;
}

int intersect_in(Line u, Line v)
{
    if (!dots_inline(v.a, u) || !dots_inline(v.b, u))
        return !same_side(u.a, u.b, v) && !same_side(v.a, v.b, u);
    return dot_online_in(u.a, v) || dot_online_in(u.b, v) || dot_online_in(v.a, u) || dot_online_in(v.b, u);
}

int opposite_side(Point p1, Point p2, Line l)
{
    return xmult(l.a, p1, l.b) * xmult(l.a, p2, l.b) < -eps;
}

int intersect_ex(Line u, Line v)
{
    return opposite_side(u.a, u.b, v) && opposite_side(v.a, v.b, u);
}

void input()
{
    scanf("%d", &m);
    n = 0;
    for (int i = 0; i < m; i++)
    {
        scanf("%lf%lf", &point[n].x, &point[n].y);
        point[n].x -= 50;
        point[n].y -= 50;
        n++;
        scanf("%lf%lf", &point[n].x, &point[n].y);
        point[n].x -= 50;
        point[n].y -= 50;
        n++;
        line[i] = Line(point[i * 2], point[i * 2 + 1]);
    }
    point[n].x = -50;
    point[n++].y = -50;
    point[n].x = 50;
    point[n++].y = 50;
    point[n].x = 50;
    point[n++].y = -50;
    point[n].x = -50;
    point[n++].y = 50;
    scanf("%lf%lf", &s.x, &s.y);
    s.x -= 50;
    s.y -= 50;
}

Point mid_point(Point &a, Point &b)
{
    return Point((a.x + b.x) / 2, (a.y + b.y) / 2);
}

void work()
{
    point[n] = point[0];
    ans = m;
    for (int i = 0; i < n; i++)
    {
        Line l(s, mid_point(point[i], point[i + 1]));
        int temp = 0;
        for (int i = 0; i < m; i++)
            if (intersect_in(l, line[i]))
                temp++;
        ans = min(ans, temp);
    }
}

int main()
{
    //freopen("t.txt", "r", stdin);
    input();
    sort(point, point + n);
//        for (int i = 0; i < n; i++)
//        printf("%f %f\n", point[i].x, point[i].y);
    work();
    printf("Number of doors = %d\n", ans + 1);
    return 0;
}