HDU 1454（Pipe）

可知最终结果一定经过一个上边缘的点和一个下边缘的点，否则可以平移直线使结果更优。

所以枚举所有上边缘点和下边缘点的点对，计算最大交点，或者通过整个管道。

#include <iostream>
#include <cmath>
#include <algorithm>
#include <iomanip>
using namespace std;
const double eps = 1e-9;
const int INF = 0x3f3f3f3f;
const int MAXN = 25;

struct Point //点
{
    double x;
    double y;
}up[MAXN], down[MAXN]; //上边缘点，下边缘点

int n;

//判断x是正数、负数、零
int judge(double x)
{
    return fabs(x) < eps ? 0 : (x > 0 ? 1 : -1);
}

//计算两个向量的叉积
double cross(Point& a, Point& b, Point& c)
{
    return (b.x - a.x) * (c.y - a.y) - (c.x - a.x) * (b.y - a.y);
}

//判断两条直线是否相交
bool isIntersect(Point& a, Point& b, Point& c, Point& d)
{
    return judge(cross(a, b, c)) * judge(cross(a, b, d)) <= 0;
}

//计算两条直线交点的横坐标
double getX(Point& a, Point& b, Point& c, Point& d)
{
    double s1 = cross(a, b, c);
    double s2 = cross(a, b, d);
    int d1 = judge(s1);
    int d2 = judge(s2);
    if (d1 * d2 != 0)
        return (s2 * c.x - s1 * d.x) / (s2 - s1);
    else if (d1 == 0)
        return c.x;
    else if (d2 == 0) 
        return d.x;
    else
        return -INF;
}

void solve()
{
    double ans = -INF;
    //枚举所有上边缘点和下边缘点的点对
    for (int i = 0; i < n; i++) //上边缘点
    {
        for (int j = 0; j < n; j++) //下边缘点
        {
            if (i == j)
                continue;
            int k;
            for (k = 0; k < n; k++)
            {
                if (!isIntersect(up[i], down[j], up[k], down[k]))
                    break;
            }
            if (k == n) //光线通过整个管道
            {
                cout << "Through all the pipe." << endl;
                return;
            }
            if (k > max(i, j))
            {
                ans = max(getX(up[i], down[j], up[k], up[k - 1]), ans);
                ans = max(getX(up[i], down[j], down[k], down[k - 1]), ans);
            }
        }
    }
    cout << fixed << setprecision(2) << ans << endl;
}

int main()
{
    while (cin >> n)
    {
        if (n == 0)
            break;
        for (int i = 0; i < n; i++)
        {
            cin >> up[i].x >> up[i].y;
            down[i].x = up[i].x;
            down[i].y = up[i].y - 1;
        }
        solve();
    }
    return 0;
}

继续加油。