POJ1039+几何+直线于线段相交

注意：

不碰任何顶点-----（上下移动）>碰一个顶点-----（绕此顶点旋转）>碰两个顶点-----（绕前后两个顶点旋转）>碰上下各一个顶点

其中：判断i，j是否被某壁阻挡，可通过判断是否与垂直线段相交判断!!!!!!!!!!

/*
几何+直线相交+求交点
*/
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
using namespace std;
const int maxn = 24;
const double inf= 99999999.0;
const double eps = 1e-8;
struct Point{
    double x,y;
}up[ maxn ],down[ maxn ];
int sig( double d ){
    if( d>eps ) return 1;
    if( d<-eps ) return -1;
    return 0;
}
double xmult( Point a,Point b,Point c ){
    return ( b.x-a.x )*( c.y-a.y )-( b.y-a.y )*( c.x-a.x );
}
bool check( Point a,Point b,Point c,Point d ){
    return (( sig(xmult(a,b,c)) )*( sig(xmult(a,b,d)) )<=0);
}
double Intersection( Point a,Point b,Point c,Point d ){
    double s1 = xmult( a,b,c );
    double s2 = xmult( a,b,d );
    int t1,t2;
    t1 = sig( s1 ),t2 = sig( s2 );
    if( t1*t2<0 ){
        return ( c.x*s2-d.x*s1 )/(s2-s1);//return x 坐标
        //(c.y*s2-d.y*s1)/(s2-s1)
    }
    if( t1*t2==0 ){
        if( t1==0 ) return c.x;
        if( t2==0 ) return d.x;
    }
    return -inf;
}
int main(){
    int n;
    while( scanf("%d",&n)==1,n ){
        for( int i=1;i<=n;i++ ){
            scanf("%lf%lf",&up[ i ].x,&up[ i ].y);
            down[ i ].x = up[ i ].x;
            down[ i ].y = up[ i ].y-1.0;
        }
        double ans = -inf;
        bool flag = false;//是否能贯通两端
        for( int i=1;i<=n;i++ ){
            for( int j=1;j<=n;j++ ){
                if( i!=j ){
                    int k;
                    for( k=1;k<=n;k++ ){
                        if( check( up[i],down[j],up[k],down[k] )==true ){}//与垂直的线段相交true
                        else break;
                    }
                    if( k>n ){
                        flag = true;
                        break;
                    }
                    else if( k>max(i,j) ){
                        double pos;
                        pos = Intersection( up[i],down[j],up[k],up[k-1] );
                        if( pos>ans ) ans = pos;
                        pos = Intersection( up[i],down[j],down[k],down[k-1] );
                        if( pos>ans ) ans = pos;
                        //因为不知道ij是和上管壁相交还是下管壁相交
                    }
                }
            }
            if( flag==true ) break;
        }
        if( flag == true ) printf("Through all the pipe.\n");
        else printf("%.2lf\n",ans);
    }
    return 0;
}