Pipe(poj1039线段相交)

题意：给你一个管道，看是光线是否从管道射出，管线没有反射，不能接触管道的边，但是能接触拐点

思路：光线穿过管道一定是经过两个拐点，一个上拐点，一个下拐点，所以枚举所有拐点，如果该直线可以穿过管道那么上方的点一定在直线的左边，下放的点一定在直线的右边，所以枚举每两个点，判断有没有上方的点在直线的右边，下方的点有没有在直线的左面，有则说明直线和管道有交点，求出交点，最后取最大值，如果线段在该线段之前的线段相交怎么说明这条直线不合法，一直枚举到最后

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>using namespace std;struct Point{    double x,y;    Point(double x = 0,double y = 0):x(x),y(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) ;}double Dot(Vector a,Vector b) { return a.x*b.x + a.y*b.y ;}double Length(Vector a) { return sqrt(Dot(a,a)) ;}double Cross(Vector a, Vector b) { return a.x*b.y - a.y*b.x ;}const double eps = 1e-8;int dcmp(double x){    if(fabs(x) < eps) return 0;    else return x < 0 ? -1 : 1;}bool operator == (Point a,Point b){    return dcmp(a.x-b.x) == 0&& dcmp(a.y-b.y) == 0;}bool operator < (Point a,Point b){    return a.x < b.x || (a.x == b.x && a.y < b.y);}bool Onsegment(Point p,Point a,Point b){    return dcmp(Cross(b-a,p-a)) == 0 && dcmp(Dot(b-p,a-p)) < 0 || (p == a) || (p == b);}bool OnLine(Point p,Point a,Point b){    return fabs(Cross(p-a,a-b)) / Length(b-a);}bool SegmentIntersection(Point a,Point b,Point c,Point d){    double d1 = Cross(b-a,c-a),d2 = Cross(b-a,d-a),d3 = Cross(d-c,a-c),d4 = Cross(d-c,b-c);    if(dcmp(d1)*dcmp(d2) < 0 || dcmp(d3)*dcmp(d4) < 0) return true;    else if(dcmp(d1) == 0 && !OnLine(c,a,b) ) return true;    else if(dcmp(d2) == 0 && !OnLine(d,a,b) ) return true;    else if(dcmp(d3) == 0 && !OnLine(a,c,d) ) return true;    else if(dcmp(d4) == 0 && !OnLine(b,c,d) ) return true;    else return false;}bool Segmentsection(Point a,Point b,Point c,Point d){    double d1 = Cross(b-a,c-a),d2 = Cross(b-a,d-a),d3 = Cross(d-c,a-c),d4 = Cross(d-c,b-c);    if(dcmp(d1)*dcmp(d2) < 0 && dcmp(d3)*dcmp(d4) < 0) return true;    else if(dcmp(d1) == 0 && Onsegment(c,a,b) ) return true;    else if(dcmp(d2) == 0 && Onsegment(d,a,b) ) return true;    else if(dcmp(d3) == 0 && Onsegment(a,c,d) ) return true;    else if(dcmp(d4) == 0 && Onsegment(b,c,d) ) return true;    else return false;}bool pointInpoly(Point p,Point a,Point b,Point c,Point d){    double d1 = Cross(b-a,p-a);    double d2 = Cross(d-c,p-c);    double d3 = Cross(c-b,p-b);    double d4 = Cross(a-d,p-d);    return dcmp(d1)*dcmp(d2) > 0 && dcmp(d3)*dcmp(d4) > 0;}Point Segment(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 Max(double a,double b){    return a > b ? a : b;}struct Line{    Point s,e;    Line(Point s = 0,Point e = 0) :s(s),e(e){}};Point up[50],down[50];int n;double ans;bool solve(Point a,Point b,int e){    int sign,i;    for( i = 0; i < n-1; i++)    {        if(dcmp( Cross(b-a,up[i]-a) ) < 0 || dcmp( Cross(b-a,up[i+1]-a) ) < 0 )        {            sign = 1;break;        }        if(  dcmp( Cross(b-a,down[i]-a) ) > 0 || dcmp( Cross(b-a,down[i+1]-a) ) > 0)        {            sign = 2;break;        }    }    if(i == n-1)        return true;    if(i < e)        return false;    Point temp;    if(sign == 1)    {        temp = Segment(a,b-a,up[i],up[i+1]-up[i]);    }    else    {        temp = Segment(a,b-a,down[i],down[i+1]-down[i]);    }    ans = Max(ans,temp.x);    return false;}int main(){    while(scanf("%d",&n) != EOF)    {        if(n == 0) break;        for(int i = 0; i < n; i++)        {            scanf("%lf%lf",&up[i].x,&up[i].y);            down[i].x = up[i].x,down[i].y = up[i].y-1;        }        ans = -9999999;        bool flag = false;              for(int i = 0; i < n && !flag; i++)        {            for(int j = i+1; j < n; j++)            {                flag = solve(up[i],down[j],j);                if(flag) break;                flag = solve(down[i],up[j],j);                if(flag) break;            }        }        if(flag)            printf("Through all the pipe.\n");        else            printf("%.2f\n",ans);    }    return 0;}