简单几何(线段与直线的位置) POJ 3304 Segments

题目传送门

题意：有若干线段，问是否存在一条直线，所有线段投影到直线上时至少有一个公共点

分析：有一个很好的解题报告：二维平面上线段与直线位置关系的判定。首先原问题可以转换为是否存在一条直线与所有线段相交，然后可以离散化枚举通过枚举端点来枚举直线，再用叉积判断直线和线段是否相交。用到了叉积

/************************************************
* Author        :Running_Time
* Created Time  :2015/10/23 星期五 17:00:08
* File Name     :POJ_3304.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 = 1e2 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double EPS = 1e-8;
struct Point    {       //点的定义
    double x, y;
    Point (double x=0, double y=0) : x (x), y (y) {}
};
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 polar_angle(Vector A)  {     //向量极角
    return atan2 (A.y, A.x);
}
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;
}
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;
}
Vector operator + (Vector A, Vector B)  {       //向量加法
    return Vector (A.x + B.x, A.y + B.y);
}
Vector operator - (Vector A, Vector B)  {       //向量减法
    return Vector (A.x - B.x, A.y - B.y);
}
Vector operator * (Vector A, double p)  {       //向量乘以标量
    return Vector (A.x * p, A.y * p);
}
Vector operator / (Vector A, double p)  {       //向量除以标量
    return Vector (A.x / p, A.y / p);
}
bool operator < (const Point &a, const Point &b)    {       //点的坐标排序
    return a.x < b.x || (a.x == b.x && a.y < b.y);
}
bool operator == (const Point &a, const Point &b)   {       //判断同一个点
    return dcmp (a.x - b.x) == 0 && dcmp (a.y - b.y) == 0;
}
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));
}
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;
}
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 point_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 dis_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 dis_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_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 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_poly(Point *p, int n)   {       //多边形面积
    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，输入点集会被修改
*/
vector<Point> convex_hull(vector<Point> &P) {
    sort (P.begin (), P.end ());
    P.erase (unique (P.begin (), P.end ()), P.end ());      //预处理，删除重复点
    int n = P.size (), m = 0;
    vector<Point> ret (n + 1);
    for (int i=0; i<n; ++i) {
        while (m > 1 && cross (ret[m-1]-ret[m-2], P[i]-ret[m-2]) < 0)   m--;
        ret[m++] = P[i];
    }
    int k = m;
    for (int i=n-2; i>=0; --i)  {
        while (m > k && cross (ret[m-1]-ret[m-2], P[i]-ret[m-2]) < 0)   m--;
        ret[m++] = P[i];
    }
    if (n > 1)  m--;
    ret.resize (m);
    return ret;
}
struct Line {
    Point s, e;
    Line () {}
    Line (Point s, Point e) : s (s), e (e) {}
};
Line L[N];
int n;

bool judge(Point a, Point b)    {
    if (a == b) return false;
    for (int i=0; i<n; ++i) {
        if (cross (a - L[i].s, b - L[i].s) * cross (a - L[i].e, b - L[i].e) > 0)   return false;
    }
    return true;
}

int main(void)    {
    int T;  scanf ("%d", &T);
    while (T--) {
        scanf ("%d", &n);
        for (int i=0; i<n; ++i) {
            L[i] = Line (read_point (), read_point ());
        }
        if (n == 1) {
            puts ("Yes!");  continue;
        }
        bool flag = false;
        for (int i=0; i<n && !flag; ++i) {
            for (int j=i+1; j<n; ++j)   {
                if (judge (L[i].s, L[j].s) || judge (L[i].s, L[j].e)
                    || judge (L[i].e, L[j].s) || judge (L[i].e, L[j].e))   {
                    flag = true;    break;
                }
            }
        }
        if (flag)   puts ("Yes!");
        else    puts ("No!");
    }

    return 0;
}