简单几何(线段相交) POJ 1066 Treasure Hunt

题目传送门

题意：从四面任意点出发，有若干障碍门，问最少要轰掉几扇门才能到达终点

分析：枚举入口点，也就是线段的两个端点，然后选取与其他线段相交点数最少的 + 1就是答案。特判一下n == 0的时候

/************************************************
* Author        :Running_Time
* Created Time  :2015/10/26 星期一 16:30:26
* File Name     :POJ_1066.cpp
 ************************************************/

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 33;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double EPS = 1e-10;
const double PI = acos (-1.0);
int dcmp(double x)  {       //三态函数，减少精度问题
    if (fabs (x) < EPS) return 0;
    else    return x < 0 ? -1 : 1;
}
struct Point    {       //点的定义
    double x, y;
    Point (double x=0, double y=0) : x (x), y (y) {}
    Point operator + (const Point &r) const {       //向量加法
        return Point (x + r.x, y + r.y);
    }
    Point operator - (const Point &r) const {       //向量减法
        return Point (x - r.x, y - r.y);
    }
    Point operator * (double p)  {       //向量乘以标量
        return Point (x * p, y * p);
    }
    Point operator / (double p)  {       //向量除以标量
        return Point (x / p, y / p);
    }
    bool operator < (const Point &r) const {       //点的坐标排序
        return x < r.x || (x == r.x && y < r.y);
    }
    bool operator == (const Point &r) const {       //判断同一个点
        return dcmp (x - r.x) == 0 && dcmp (y - r.y) == 0;
    }
};
typedef Point Vector;       //向量的定义
Point read_point(void)   {      //点的读入
    double x, y;
    scanf ("%lf%lf", &x, &y);
    return Point (x, y);
}
double dot(Vector A, Vector B)  {       //向量点积
    return A.x * B.x + A.y * B.y;
}
double cross(Vector A, Vector B)    {       //向量叉积
    return A.x * B.y - A.y * B.x;
}
double polar_angle(Vector A)  {     //向量极角
    return atan2 (A.y, A.x);
}
double length(Vector A) {       //向量长度，点积
    return sqrt (dot (A, A));
}
double angle(Vector A, Vector B)    {       //向量转角，逆时针，点积
    return acos (dot (A, B) / length (A) / length (B));
}
Vector rotate(Vector A, double rad) {       //向量旋转，逆时针
    return Vector (A.x * cos (rad) - A.y * sin (rad), A.x * sin (rad) + A.y * cos (rad));
}
Vector nomal(Vector A)  {       //向量的单位法向量
    double len = length (A);
    return Vector (-A.y / len, A.x / len);
}
Point line_line_inter(Point p, Vector V, Point q, Vector W)    {        //两直线交点，参数方程
    Vector U = p - q;
    double t = cross (W, U) / cross (V, W);
    return p + V * t;
}
double point_to_line(Point p, Point a, Point b)   {       //点到直线的距离，两点式
    Vector V1 = b - a, V2 = p - a;
    return fabs (cross (V1, V2)) / length (V1);
}
double point_to_seg(Point p, Point a, Point b)    {       //点到线段的距离，两点式
    if (a == b) return length (p - a);
    Vector V1 = b - a, V2 = p - a, V3 = p - b;
    if (dcmp (dot (V1, V2)) < 0)    return length (V2);
    else if (dcmp (dot (V1, V3)) > 0)   return length (V3);
    else    return fabs (cross (V1, V2)) / length (V1);
}
Point point_line_proj(Point p, Point a, Point b)   {     //点在直线上的投影，两点式
    Vector V = b - a;
    return a + V * (dot (V, p - a) / dot (V, V));
}
bool can_inter(Point a1, Point a2, Point b1, Point b2)  {       //判断线段相交，两点式
    double c1 = cross (a2 - a1, b1 - a1), c2 = cross (a2 - a1, b2 - a1),
           c3 = cross (b2 - b1, a1 - b1), c4 = cross (b2 - b1, a2 - b1);
    return dcmp (c1) * dcmp (c2) < 0 && dcmp (c3) * dcmp (c4) < 0;
}
bool on_seg(Point p, Point a1, Point a2)    {       //判断点在线段上，两点式
    return dcmp (cross (a1 - p, a2 - p)) == 0 && dcmp (dot (a1 - p, a2 - p)) < 0;
}
double area_triangle(Point a, Point b, Point c) {       //三角形面积，叉积
    return fabs (cross (b - a, c - a)) / 2.0;
}
double area_poly(Point *p, int n)   {       //多边形面积，叉积
    double ret = 0;
    for (int i=1; i<n-1; ++i)   {
        ret += fabs (cross (p[i] - p[0], p[i+1] - p[0]));
    }
    return ret / 2;
}
/*
    点集凸包
*/
vector<Point> convex_hull(vector<Point> &P) {
    sort (P.begin (), P.end ());
    int n = P.size (), k = 0;
    vector<Point> ret (n * 2);
    for (int i=0; i<n; ++i) {
        while (k > 1 && cross (ret[k-1] - ret[k-2], P[i] - ret[k-1]) <= 0)  k--;
        ret[k++] = P[i];
    }
    for (int i=n-2, t=k; i>=0; --i)  {
        while (k > t && cross (ret[k-1] - ret[k-2], P[i] - ret[k-1]) <= 0)  k--;
        ret[k++] = P[i];
    }
    ret.resize (k-1);
    return ret;
}

struct Circle   {
    Point c;
    double r;
    Circle () {}
    Circle (Point c, double r) : c (c), r (r) {}
    Point point(double a)   {
        return Point (c.x + cos (a) * r, c.y + sin (a) * r);
    }
};
struct Line {
    Point p;
    Vector v;
    double r;
    Line () {}
    Line (const Point &p, const Vector &v) : p (p), v (v) {
        r = polar_angle (v);
    }
    Point point(double a)   {
        return p + v * a;
    }
};

Point s[N], e[N];
Point p;
int cnt1[N], cnt2[N];

int main(void)    {
    int n;  scanf ("%d", &n);
    for (int i=1; i<=n; ++i)    {
        s[i] = read_point ();
        e[i] = read_point ();
    }
    p = read_point ();
    if (!n) {
        printf ("Number of doors = %d\n", 1);   return 0;
    }
    int ans = INF;
    for (int i=1; i<=n; ++i)    {
        for (int j=1; j<=n; ++j)    {
            if (i == j) continue;
            if (can_inter (p, s[i], s[j], e[j]))    {
                cnt1[i]++;
            }
            if (can_inter (p, e[i], s[j], e[j]))    {
                cnt2[i]++;
            }
        }
        ans = min (ans, min (cnt1[i], cnt2[i]));
    }

    printf ("Number of doors = %d\n", ans + 1);

   //cout << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n";

    return 0;
}