POJ 3304 Segments(计算几何)

意甲冠军：给出的一些段的。问：能否找到一条直线，通过所有的行

思维：假设一条直线的存在，所以必须有该过两点的线，然后列举两点，然后推断是否存在与所有的行的交点可以是

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));
}

//推断3点共线
bool LineCoincide(Point p1, Point p2, Point p3) {
    return dcmp(Cross(p2 - p1, p3 - p1)) == 0;
}

//推断向量平行
bool LineParallel(Vector v, Vector w) {
    return Cross(v, w) == 0;
}

//推断向量垂直
bool LineVertical(Vector v, Vector w) {
    return Dot(v, w) == 0;
}

//计算两直线交点，平行，重合要先推断
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;
}

//推断点在线段上, 不包括端点
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;
}

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