简单几何(线段与直线的位置) POJ 3304 Segments-白红宇

简单几何(线段与直线的位置) POJ 3304 Segments

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发布时间：2019-06-27

本文共 4308 字，大约阅读时间需要 14 分钟。

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：。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 
     
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                      using namespace std;#define lson l, mid, rt << 1#define rson mid + 1, r, rt << 1 | 1typedef 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
                      
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                         &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
                        
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                           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
                          
                            0) return false; } return true;}int main(void) { int T; scanf ("%d", &T); while (T--) { scanf ("%d", &n); for (int i=0; i