POJ 1039 直线和线段相交

题意：

题意很好理解，从左边射过来的光线，最远能经过管道到右边多少距离。

分析：

光线一定经过一个上端点和一个下端点，这一点很容易想到。然后枚举上下端点即可

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#define eps 1e-8
#define INF 1e9
#define OK sgn(tmp-lx1)>0 && sgn(tmp-lx2<0)
using namespace std;

typedef struct Point
{
    double x,y;
    Point() {};
    Point(double xx,double yy)
    {
        x=xx;
        y=yy;
    }
} Vector;

struct Line
{
    Point p,q;
    Line() {};
    Line(Point pp,Point qq)
    {
        p=pp;
        q=qq;
    }
};

int sgn(double x)
{
    if(fabs(x)<eps) return 0;
    return x<0? -1:1;
}

double crs_prdct(Vector a,Vector b)
{
    return a.x*b.y-b.x*a.y;
}

Vector operator - (Point a,Point b)
{
    return Vector(a.x-b.x,a.y-b.y);
}

double cross_x(Point p1,Point p2,Point p3,Point p4)
{
    double k1=(p1.y-p2.y)/(p1.x-p2.x);
    double k2=(p3.y-p4.y)/(p3.x-p4.x);
    return (k1*p1.x-k2*p3.x-p1.y+p3.y)/(k1-k2);
}

double get_y(Point p,Point q,double x)
{
    return (p.y-q.y)/(p.x-q.x)*(x-p.x)+p.y;
}

//判断直线和线段相交
bool Seg_inter_line(Line l1,Line l2) //判断直线l1和线段l2是否相交,<0是把交于线段端点处看做不相交
{
    return sgn(crs_prdct(l2.p-l1.q,l1.p-l1.q))*sgn(crs_prdct(l2.q-l1.q,l1.p-l1.q)) < 0;
}

const int maxn=30;
Point Up[maxn],Dw[maxn];

int main()
{
//    freopen("in.txt","r",stdin);
    int n;
    while(scanf("%d",&n),n)
    {
        double x,y;
        for(int i=0; i<n; i++)
        {
            scanf("%lf%lf",&x,&y);
            Up[i]=Point(x,y);
            Dw[i]=Point(x,y-1);
        }
        double ans=Up[0].x;
        for(int i=0; i<n; i++)
        {
            for(int j=0; j<n; j++)
            {
                if(i==j) continue;
                for(int k=0; k<n-1; k++)
                {
                    if(Seg_inter_line(Line(Up[i],Dw[j]),Line(Up[k],Up[k+1])))
                    {
                        ans=max(ans,cross_x(Up[i],Dw[j],Up[k],Up[k+1]));
                        break;
                    }
                    if(Seg_inter_line(Line(Up[i],Dw[j]),Line(Dw[k],Dw[k+1])))
                    {
                        ans=max(ans,cross_x(Up[i],Dw[j],Dw[k],Dw[k+1]));
                        break;
                    }
                    double t=get_y(Up[i],Dw[j],(Up[k].x+Up[k+1].x)/2);
                    if(t>(Up[k].y+Up[k+1].y)/2 || t<(Dw[k].y+Dw[k+1].y)/2)
                    {
                        ans=max(ans,Up[k].x);
                        break;
                    }
                    if(k==n-2) ans=Up[n-1].x;
                }
            }
        }
        if(sgn(ans-Up[n-1].x)==0) printf("Through all the pipe.
");
        else printf("%.2f
",ans);
    }
    return 0;
}