题意：给定一些线段，问能否找出一条直线，，穿过所有线段
思路：如果存在一条直线，那么必然有一条直线是过已有的两点，那么就枚举两点，然后去判断是否跟所有线段有交点即可
代码：
#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>using namespace std;struct Point {double x, y;Point() {}Point(double x, double y) {this->x = x;this->y = y;}void read() {scanf("%lf%lf", &x, &y);}};typedef Point Vector;Vector operator + (Vector A, Vector B) {return Vector(A.x + B.x, A.y + B.y);}Vector operator – (Vector A, Vector B) {return Vector(A.x – B.x, A.y – B.y);}Vector operator * (Vector A, double p) {return Vector(A.x * p, A.y * p);}Vector operator / (Vector A, double p) {return Vector(A.x / p, A.y / p);}bool operator < (const Point& a, const Point& b) {return a.x < b.x || (a.x == b.x && a.y < b.y);}const double eps = 1e-8;int dcmp(double x) {if (fabs(x) < eps) return 0;else return x < 0 ? -1 : 1;}bool operator == (const Point& a, const Point& b) {return dcmp(a.x – b.x) == 0 && dcmp(a.y – b.y) == 0;}double Dot(Vector A, Vector B) {return A.x * B.x + A.y * B.y;} //点积double Length(Vector A) {return sqrt(Dot(A, A));} //向量的模double Angle(Vector A, Vector B) {return acos(Dot(A, B) / Length(A) / Length(B));} //向量夹角double Cross(Vector A, Vector B) {return A.x * B.y – A.y * B.x;} //叉积double Area2(Point A, Point B, Point C) {return Cross(B – A, C – A);} //有向面积//向量旋转Vector Rotate(Vector A, double rad) {return Vector(A.x * cos(rad) – A.y * sin(rad), A.x * sin(rad) + A.y * cos(rad));}//判断3点共线bool LineCoincide(Point p1, Point p2, Point p3) {return dcmp(Cross(p2 – p1, p3 – p1)) == 0;}//判断向量平行bool LineParallel(Vector v, Vector w) {return Cross(v, w) == 0;}//判断向量垂直bool LineVertical(Vector v, Vector w) {return Dot(v, w) == 0;}//计算两直线交点，平行，重合要先判断Point GetLineIntersection(Point P, Vector v, Point Q, Vector w) {Vector u = P – Q;double t = Cross(w, u) / Cross(v, w);return P + v * t;}//点到直线距离double DistanceToLine(Point P, Point A, Point B) {Vector v1 = B – A, v2 = P – A;return fabs(Cross(v1, v2)) / Length(v1);}//点到线段距离double DistanceToSegment(Point P, Point A, Point B) {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);}//点在直线上的投影点Point GetLineProjection(Point P, Point A, Point B) {Vector v = B – A;return A + v * (Dot(v, P – A) / Dot(v, v));}//线段相交判定(规范相交)bool SegmentProperIntersection(Point a1, Point a2, Point b1, Point b2) {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);//dcmp(c1) * dcmp(c2) == 0 || dcmp(c3) * dcmp(c4) == 0为不规范相交return dcmp(c1) * dcmp(c2) <= 0;// && dcmp(c3) * dcmp(c4) <= 0;}//判断点在线段上, 不包含端点bool OnSegment(Point p, Point a1, Point a2) {return dcmp(Cross(a1 – p, a2 – p)) == 0 && dcmp(Dot(a1 – p, a2 – p)) < 0;}//n边形的面积double PolygonArea(Point *p, int n) {double area = 0;for (int i = 1; i < n – 1; i++)area += Cross(p[i] – p[0], p[i + 1] – p[0]);return area / 2;}const int N = 105;int t, n;struct Line {Point a, b;void read() {a.read();b.read();}} line[N];bool judge(Point a, Point b) {if (dcmp(a.x – b.x) == 0 && dcmp(a.y – b.y) == 0) return false;for (int i = 0; i < n; i++)if (!SegmentProperIntersection(a, b, line[i].a, line[i].b)) return false;return true;}bool gao() {for (int i = 0; i < n; i++) {if (judge(line[i].a, line[i].b)) return true;for (int j = 0; j < i; j++) {if (judge(line[i].a, line[j].a)) return true;if (judge(line[i].a, line[j].b)) return true;if (judge(line[i].b, line[j].a)) return true;if (judge(line[i].b, line[j].b)) return true;}}return false;}int main() {scanf("%d", &t);while (t–) {scanf("%d", &n);for (int i = 0; i < n; i++)line[i].read();if (gao()) printf("Yes!\n");else printf("No!\n");}return 0;}
未经一番寒彻骨，焉得梅花扑鼻香
