POJ 3304 (計算幾何)

題目鏈接：點擊這裏

題意：給出n個線段，能否找到一個直線使得所有線段在直線上的投影至少有一個交點。

可以轉化爲找到與n個線段都相交的直線，那麼所有的直線就是與這條垂直的直線。所以直接枚舉任意兩個直線的任意兩個端點。
trick:當兩個點的距離小於精度時認爲他們是同一個點直接跳過。

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

#define maxn 105
const double eps = 1e-8;
const double INF = 1e20;
const double pi = acos (-1.0);

int dcmp (double x) {
    if (fabs (x) < eps) return 0;
    return (x < 0 ? -1 : 1);
}
inline double sqr (double x) {return x*x;}

//*************點
struct Point {
    double x, y;
    Point (double _x = 0, double _y = 0):x(_x), y(_y) {}
    void input () {scanf ("%lf%lf", &x, &y);}
    void output () {printf ("%.2f %.2f\n", x, y);}
    bool operator == (const Point &b) const {
        return (dcmp (x-b.x) == 0 && dcmp (y-b.y) == 0);
    }
    bool operator < (const Point &b) const {
        return (dcmp (x-b.x) == 0 ? dcmp (y-b.y) < 0 : x < b.x);
    }
    Point operator + (const Point &b) const {
        return Point (x+b.x, y+b.y);
    }
    Point operator - (const Point &b) const {
        return Point (x-b.x, y-b.y);
    }
    Point operator * (double a) {
        return Point (x*a, y*a);
    }
    Point operator / (double a) {
        return Point (x/a, y/a);
    }
    double len2 () {//返回長度的平方
        return sqr (x) + sqr (y);
    }
    double len () {//返回長度
        return sqrt (len2 ());
    }
    Point change_len (double r) {//轉化爲長度爲r的向量
        double l = len ();
        if (dcmp (l) == 0) return *this;//零向量返回自身
        r /= l;
        return Point (x*r, y*r);
    }
    Point rotate_left () {//順時針旋轉90度
        return Point (-y, x);
    }
    Point rotate_right () {//逆時針旋轉90度
        return Point (y, -x);
    }
    Point rotate (Point p, double ang) {//繞點p逆時針旋轉ang
        Point v = (*this)-p;
        double c = cos (ang), s = sin (ang);
        return Point (p.x + v.x*c - v.y*s, p.y + v.x*s + v.y*c);
    }
    Point normal () {//單位法向量
        double l = len ();
        return Point (-y/l, x/l);
    }
};

double cross (Point a, Point b) {//叉積
    return a.x*b.y-a.y*b.x;
}
double dot (Point a, Point b) {//點積
    return a.x*b.x + a.y*b.y;
}
double dis (Point a, Point b) {//兩個點的距離
    Point p = b-a; return p.len ();
}
double degree_rad (double ang) {//角度轉化爲弧度   
    return ang/180*pi;
}
double rad_degree (double rad) {//弧度轉化爲角度
    return rad/pi*180;
}
double rad (Point a, Point b) {//兩個向量的夾角
    return fabs (atan2 (fabs (cross (a, b)), dot (a, b)) );
}
bool parallel (Point a, Point b) {//向量平行
    double p = rad (a, b);
    return dcmp (p) == 0 || dcmp (p-pi) == 0;
}

//************直線 線段
struct Line {
    Point s, e;//直線的兩個點
    double k;//極角 範圍[-pi,pi]
    Line () {}
    Line (Point _s, Point _e) {
        s = _s, e = _e;
        k = atan2 (e.y - s.y,e.x - s.x);
    }
    //ax+by+c = 0
    Line (double a, double b, double c) {
        if (dcmp (a) == 0) {
            s = Point (0, -c/b);
            e = Point (1, -c/b);
        }
        else if (dcmp (b) == 0) {
            s = Point (-c/a, 0);
            e = Point (-c/a, 1);
        }
        else {
            s = Point (0, -c/b);
            e = Point (1, (-c-a)/b);
        }
        get_angle ();
    }
    //一個點和傾斜角確定直線
    Line (Point p, double ang) {
        k = ang;
        s = p;
        if (dcmp (ang-pi/2) == 0) {
            e = s + Point (0, 1);
        }
        else
            e = s + Point (1, tan (ang));
    }
    void input () {
        s.input ();
        e.input ();
    }
    void output () {
        printf ("%.2f,%.2f %.2f,%.2f\n", s.x, s.y, e.x, e.y);
    }
    void adjust () {
        if (e < s) swap (e, s);
    }
    double length () {//求線段長度
        return dis (s, e);
    }
    void get_angle () {
        k = atan2 (e.y - s.y,e.x - s.x);
    }
    double angle () {//直線的傾斜角
        if (dcmp (k) < 0) k += pi;
        if (dcmp (k-pi) == 0) k -= pi;
        return k;
    }
    Point operator &(const Line &b)const {//直線的交點(保證存在)
        Point res = s;
        double t = (cross (s - b.s, b.s - b.e))/cross (s - e, b.s - b.e); 
        res.x += (e.x - s.x)*t;
        res.y += (e.y - s.y)*t;
        return res;
    }
};

int relation (Point p, Line l) {//點和直線的關係
    //1:在左側 2:在右側 3:在直線上
    int c = dcmp (cross (p-l.s, l.e-l.s));
    if (c < 0) return 1;
    else if (c > 0) return 2;
    else return 3;
}

bool point_on_halfline (Point p, Line l) {//判斷點在射線上
    int id = relation (p, l);
    if (id != 3) return 0;
    return dcmp (dot (p-l.s, l.e-l.s)) >= 0;
}

bool point_on_seg (Point p, Line l) {//判斷點在線段上
    return dcmp (cross (p-l.s, l.e-l.s)) == 0 &&
    dcmp (dot (p-l.s, p-l.e) <= 0);
    //如果忽略端點交點改成小於號就好了
}

bool parallel (Line a, Line b) {//直線平行
    return parallel (a.e-a.s, b.e-b.s);
}

int seg_cross_seg (Line a, Line v) {//線段相交判斷
    //2:規範相交 1:不規範相交 0:不相交
    int d1 = dcmp (cross (a.e-a.s, v.s-a.s));
    int d2 = dcmp (cross (a.e-a.s, v.e-a.s));
    int d3 = dcmp (cross (v.e-v.s, a.s-v.s));
    int d4 = dcmp (cross (v.e-v.s, a.e-v.s));
    if ((d1^d2) == -2 && (d3^d4) == -2) return 2;
    return (d1 == 0 && dcmp (dot (v.s-a.s, v.s-a.e)) <= 0) ||
        (d2 == 0 && dcmp (dot (v.e-a.s, v.e-a.e)) <= 0) ||
        (d3 == 0 && dcmp (dot (a.s-v.s, a.s-v.e)) <= 0) ||
        (d4 == 0 && dcmp (dot (a.e-v.s, a.e-v.e)) <= 0);
}

int line_cross_seg (Line a, Line v) {//直線和線段相交判斷 a直線v線段
    //2:規範相交 1:非規範相交 0:不相交
    int d1 = dcmp (cross (a.e-a.s, v.s-a.s));
    int d2 = dcmp (cross (a.e-a.s, v.e-a.s));
    if ((d1^d2) == -2) return 2;
    return (d1 == 0 || d2 == 0);
}

Line l[maxn];
int n;

bool ok (int i, int j, Point a, Point b) {
    if (dis (a, b) <= eps) return 0;
    Line tmp = Line (a, b);
    for (int k = 1; k <= n; k++) if (k != i && k != j) {
        if (line_cross_seg (tmp, l[k]) == 0) return 0;
    }
    return 1;
}

int main () {
    int t;
    cin >> t;
    while (t--) {
        cin >> n;
        for (int i = 1; i <= n; i++) l[i].input ();
        if (n == 1 || n == 2) {
            printf ("Yes!\n");
            continue;
        }
        for (int i = 1; i <= n; i++) {
            for (int j = i+1; j <= n; j++) {
                if (ok (i, j, l[i].s, l[j].s) || ok (i, j, l[i].s, l[j].e) || 
                    ok (i, j, l[i].e, l[j].e) || ok (i, j, l[i].e, l[j].s)) {
                    printf ("Yes!\n");
                    goto out;
                }
            }
        }
        printf ("No!\n");
        out: ;
    }
    return 0;
}