poj 1066 Treasure Hunt 计算几何计数

题目地址：poj1066

只能从中点进入--姑且理解为要进入另一个区域，一定只能“边相邻”。

关于穿过最小边的方案--把目标点和每一个边上的点连线，相交，交点数+1 就是可能的结果之一。

额，最小性的只是感觉上的，还没有严格的证明，以后想到了再补充吧。

代码：

#include<iostream>#include<cmath>#include<cstdio>#include<algorithm>#include<cstring>#include<map>#include<vector>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;}double min(double x,double y){    return x<y?x:y;    }double max(double x,double y){    return x>y?x:y;    }Point  p[100];Point  a[35],b[35];int main(){    int n;    cin>>n;    for(int i=0;i<n;i++)    {        a[i]=read_point();        b[i]=read_point();            }        int cnt=0;    for(int i=0;i<n;i++)    {        p[cnt++]=a[i];        p[cnt++]=b[i];    }        p[cnt++]=Point(0,0);    p[cnt++]=Point(0,100);    p[cnt++]=Point(100,0);    p[cnt++]=Point(100,100);           Point  target;    target=read_point();        int ans;        bool first=1;        for(int i=0;i<cnt;i++)    {        int   count=0;            for(int j=0;j<n;j++)        {            if(SegmentProperIntersection(p[i], target, a[j], b[j]))                count++;                    }                if(first)   ans=count;        if(first)   first=0;        else        {            ans=min(ans,count);                    }            }        printf("Number of doors = %d\n",ans+1);    }