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

 

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/************************************************ * 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; }