poj 3304 Segments 是否存在一条穿过所有线段的直线

题目地址：poj3304

n很小，只要实现就可以了，暴力。

首先转化命题，原命题等价为，存在一条穿过所有线段的直线，

这个命题进一步等价为，取出所有点集中的点对，存在一组点对所在的直线和所有线段相交。

值得注意的是，要特判两个点相等这个特殊情况啊。

额，计算几何一定要注意“重合”二字。

代码：

#include<iostream>#include<cmath>#include<cstdio>#include<algorithm>const double eps=1e-8;using namespace std;struct Point{    double x;    double y;    Point(double x=0,double y=0):x(x),y(y){}    void operator<<(Point &A) {cout<<A.x<<' '<<A.y<<endl;}};int dcmp(double x)  {return (x>eps)-(x<-eps); }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);}// ps  coutostream &operator<<(ostream & out,Point & P) { out<<P.x<<' '<<P.y<<endl; return out;}bool  operator< (const Point &A,const Point &B) { return A.x<B.x||(A.x==B.x&&A.y<B.y); }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;}double  Cross(Vector A,Vector B)  {return A.x*B.y-B.x*A.y; }double  Length(Vector A)  { return sqrt(Dot(A, A));}double  Angle(Vector A,Vector B) {return acos(Dot(A,B)/Length(A)/Length(B));}double  Area2(Point A,Point B,Point C ) {return Cross(B-A, C-A);}Vector Rotate(Vector A,double rad) { return Vector(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad));}Vector Normal(Vector A) {double L=Length(A);return Vector(-A.y/L,A.x/L);}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;    }double DistanceToLine(Point P,Point A,Point B){    Vector v1=P-A; Vector v2=B-A;    return fabs(Cross(v1,v2))/Length(v2);    }double DistanceToSegment(Point P,Point A,Point B){    if(A==B)  return Length(P-A);        Vector v1=B-A;    Vector v2=P-A;    Vector v3=P-B;        if(dcmp(Dot(v1,v2))==-1)    return  Length(v2);    else if(Dot(v1,v3)>0)    return Length(v3);        else return DistanceToLine(P, A, B);    }Point GetLineProjection(Point P,Point A,Point B){    Vector v=B-A;    Vector v1=P-A;    double t=Dot(v,v1)/Dot(v,v);        return  A+v*t;}bool  SegmentProperIntersection(Point a1,Point a2,Point b1,Point b2){    double c1=Cross(b1-a1, a2-a1);    double c2=Cross(b2-a1, a2-a1);    double c3=Cross(a1-b1, b2-b1);    double c4=Cross(a2-b1, b2-b1);            return dcmp(c1)*dcmp(c2)<0&&dcmp(c3)*dcmp(c4)<0 ;    }bool  OnSegment(Point P,Point A,Point B){    return dcmp(Cross(P-A, P-B))==0&&dcmp(Dot(P-A,P-B))<=0;}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;    }Point  read_point(){    Point P;    scanf("%lf%lf",&P.x,&P.y);    return  P;}Point a[105],b[105],c[310];int n;bool  SegmentLineIntersection(Point a1,Point a2,Point b1,Point b2)   // Line a1,a2{    double c1=Cross(b1-a1, a2-a1);    double c2=Cross(b2-a1, a2-a1);        return dcmp(c1)*dcmp(c2)<=0 ;    }bool fine(Point A,Point B){    bool  ans=1;        if(A==B)  return 0;        for(int i=0;i<n;i++)    {        if(!SegmentLineIntersection(A, B, a[i],b[i]))        {            ans=0;            break;        }    }        return ans;    }int main(){    int T;    cin>>T;       while(T--)    {        cin>>n;                int cnt=0;        for(int i=0;i<n;i++)        {            a[i]=read_point();            b[i]=read_point();            c[cnt++]=a[i];            c[cnt++]=b[i];        }                bool ok=0;        bool out_br=0;                for(int i=0;i<2*n;i++)        {             for(int j=i+1;j<2*n;j++)            {                if(fine(c[i],c[j]))                {                    ok=1;                    out_br=1;                    break;                }            }                        if(out_br) break;        }                        if(ok) cout<<"Yes!"<<endl;        else cout<<"No!"<<endl;    }    }