学步园

枚举任意两个点作为直线。

看是否能穿越整个管道即可。

判断相交使用叉积。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <map>
#include <sstream>
#include <queue>
#include <vector>
#define MAXN 100005
#define MAXM 211111
#define eps 1e-8
#define INF 50000001
using namespace std;
inline int dblcmp(double d)
{
    if(fabs(d) < eps) return 0;
    return d > eps ? 1 : -1;
}
struct point
{
    double x, y;
    point(){}
    point(double _x, double _y): x(_x), y(_y) {}
    void input()
    {
        scanf("%lf%lf", &x, &y);
    }
    bool operator ==(point a)const
    {
        return dblcmp(a.x - x) == 0 && dblcmp(a.y - y) == 0;
    }
    point sub(point p)
    {
        return point(x - p.x, y - p.y);
    }
    double dot(point p)
    {
        return x * p.x + y * p.y;
    }
    double det(point p)
    {
        return x * p.y - y * p.x;
    }
    double distance(point p)
    {
        return hypot(x - p.x, y - p.y);
    }
}a[25], b[25];
struct line
{
    point a, b;
    line(){}
    line(point _a, point _b){ a = _a; b = _b;}
    void input()
    {
        a.input();
        b.input();
    }
    int segcrossseg(line v)
    {
        int d1 = dblcmp(b.sub(a).det(v.a.sub(a)));
        int d2 = dblcmp(b.sub(a).det(v.b.sub(a)));
        int d3 = dblcmp(v.b.sub(v.a).det(a.sub(v.a)));
        int d4 = dblcmp(v.b.sub(v.a).det(b.sub(v.a)));
        if ((d1 ^ d2) == -2 && (d3 ^ d4) == -2)return 2;
        return (d1 == 0 && dblcmp(v.a.sub(a).dot(v.a.sub(b))) <= 0||
                d2 == 0 && dblcmp(v.b.sub(a).dot(v.b.sub(b))) <= 0||
                d3 == 0 && dblcmp(a.sub(v.a).dot(a.sub(v.b))) <= 0||
                d4 == 0 && dblcmp(b.sub(v.a).dot(b.sub(v.b))) <= 0);
    }
    int linecrossseg(line v)//v is seg
    {
        int d1 = dblcmp(b.sub(a).det(v.a.sub(a)));
        int d2 = dblcmp(b.sub(a).det(v.b.sub(a)));
        if ((d1 ^ d2) == -2) return 2;
        return (d1 == 0 || d2 == 0);
    }

    point crosspoint(line v)
    {
        double a1 = v.b.sub(v.a).det(a.sub(v.a));
        double a2 = v.b.sub(v.a).det(b.sub(v.a));
        return point((a.x * a2 - b.x * a1) / (a2 - a1), (a.y * a2 - b.y * a1) / (a2 - a1));
    }

};
int n;
double ans;
int gao(line x, int id)
{
    int ind = -1;
    int kind = -1;
    for(int i = 0; i < n - 1; i ++)
    {
        if(dblcmp(x.b.sub(x.a).det(a[i].sub(x.a))) < 0 || dblcmp(x.b.sub(x.a).det(a[i + 1].sub(x.a))) < 0)
        {
            ind = i;
            kind = 1;
            break;
        }
        if(dblcmp(x.b.sub(x.a).det(b[i].sub(x.a))) > 0 || dblcmp(x.b.sub(x.a).det(b[i + 1].sub(x.a))) > 0)
        {
            ind = i;
            kind = 2;
            break;
        }
    }
    if(ind != -1 && ind < id) return 0;
    if(ind == -1) return 1;
    line y;
    if(kind == 1) y = line(a[ind], a[ind + 1]);
    else y = line(b[ind], b[ind + 1]);
    point c = x.crosspoint(y);
    ans = max(ans, c.x);
    return 0;
}
int main()
{
    while(scanf("%d", &n) != EOF && n)
    {
        for(int i = 0; i < n; i++)
        {
            a[i].input();
            b[i].x = a[i].x;
            b[i].y = a[i].y - 1;
        }
        ans = a[0].x;
        int flag = 0;
        for(int i = 0; i < n; i++)
            for(int j = i + 1; j < n; j++)
            {
                line x = line(a[i], b[j]);
                flag |= gao(x, j);
                x = line(b[i], a[j]);
                flag |= gao(x, j);
                if(flag) break;
            }
        if(flag) puts("Through all the pipe.");
        else printf("%.2f\n", ans);
    }
    return 0;
}