简单几何(直线与线段相交) POJ 1039 Pipe

题目传送门

题意：一根管道，有光源从入口发射，问光源最远到达的地方。

分析：黑书上的例题，解法是枚举任意的一个上顶点和一个下顶点(优化后)，组成直线，如果直线与所有竖直线段有交点，则表示能穿过管道。

/************************************************
* Author        :Running_Time
* Created Time  :2015/10/31 星期六 10:28:12
* File Name     :POJ_1039.cpp
 ************************************************/

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double EPS = 1e-10;
const double PI = acos (-1.0);
int dcmp(double x)  {       //三态函数，减少精度问题
    if (fabs (x) < EPS) return 0;
    else    return x < 0 ? -1 : 1;
}
struct Point    {       //点的定义
    double x, y;
    Point () {}
    Point (double x, double y) : x (x), y (y) {}
    Point operator + (const Point &r) const {       //向量加法
        return Point (x + r.x, y + r.y);
    }
    Point operator - (const Point &r) const {       //向量减法
        return Point (x - r.x, y - r.y);
    }
    Point operator * (double p) const {       //向量乘以标量
        return Point (x * p, y * p);
    }
    Point operator / (double p) const {       //向量除以标量
        return Point (x / p, y / p);
    }
    bool operator < (const Point &r) const {       //点的坐标排序
        return x < r.x || (x == r.x && y < r.y);
    }
    bool operator == (const Point &r) const {       //判断同一个点
        return dcmp (x - r.x) == 0 && dcmp (y - r.y) == 0;
    }
};
typedef Point Vector;       //向量的定义
Point read_point(void)   {      //点的读入
    double x, y;
    scanf ("%lf%lf", &x, &y);
    return Point (x, y);
}
double dot(Vector A, Vector B)  {       //向量点积
    return A.x * B.x + A.y * B.y;
}
double cross(Vector A, Vector B)    {       //向量叉积
    return A.x * B.y - A.y * B.x;
}
double polar_angle(Vector A)  {     //向量极角
    return atan2 (A.y, A.x);
}
double length(Vector A) {       //向量长度，点积
    return sqrt (dot (A, A));
}
double angle(Vector A, Vector B)    {       //向量转角，逆时针，点积
    return acos (dot (A, B) / length (A) / length (B));
}
Vector rotate(Vector A, double rad) {       //向量旋转，逆时针
    return Vector (A.x * cos (rad) - A.y * sin (rad), A.x * sin (rad) + A.y * cos (rad));
}
Vector nomal(Vector A)  {       //向量的单位法向量
    double len = length (A);
    return Vector (-A.y / len, A.x / len);
}
Point line_line_inter(Point p, Vector V, Point q, Vector W)    {        //两直线交点，参数方程
    Vector U = p - q;
    double t = cross (W, U) / cross (V, W);
    return p + V * t;
}
double point_to_line(Point p, Point a, Point b)   {       //点到直线的距离，两点式
    Vector V1 = b - a, V2 = p - a;
    return fabs (cross (V1, V2)) / length (V1);
}
double point_to_seg(Point p, Point a, Point b)    {       //点到线段的距离，两点式
    if (a == b) return length (p - a);
    Vector V1 = b - a, V2 = p - a, V3 = p - b;
    if (dcmp (dot (V1, V2)) < 0)    return length (V2);
    else if (dcmp (dot (V1, V3)) > 0)   return length (V3);
    else    return fabs (cross (V1, V2)) / length (V1);
}
Point point_line_proj(Point p, Point a, Point b)   {     //点在直线上的投影，两点式
    Vector V = b - a;
    return a + V * (dot (V, p - a) / dot (V, V));
}
bool can_seg_seg_inter(Point a1, Point a2, Point b1, Point b2)  {       //判断线段相交，两点式
    double c1 = cross (a2 - a1, b1 - a1), c2 = cross (a2 - a1, b2 - a1),
           c3 = cross (b2 - b1, a1 - b1), c4 = cross (b2 - b1, a2 - b1);
    return dcmp (c1) * dcmp (c2) < 0 && dcmp (c3) * dcmp (c4) < 0;
}
bool can_line_seg_inter(Point a1, Point a2, Point b1, Point b2)    {
    double c1 = cross (a2 - a1, b1 - a1), c2 = cross (a2 - a1, b2 - a1);
    return dcmp (c1 * c2) <= 0;
}
bool on_seg(Point p, Point a1, Point a2)    {       //判断点在线段上，两点式
    return dcmp (cross (a1 - p, a2 - p)) == 0 && dcmp (dot (a1 - p, a2 - p)) < 0;
}
double area_triangle(Point a, Point b, Point c) {       //三角形面积，叉积
    return fabs (cross (b - a, c - a)) / 2.0;
}
double area_poly(Point *p, int n)   {       //多边形面积，叉积
    double ret = 0;
    for (int i=1; i<n-1; ++i)   {
        ret += fabs (cross (p[i] - p[0], p[i+1] - p[0]));
    }
    return ret / 2;
}
/*
    点集凸包，输入点的集合，返回凸包点的集合。
	如果不希望在凸包的边上有输入点，把两个 <= 改成 <
*/
vector<Point> convex_hull(vector<Point> ps) {
    sort (ps.begin (), ps.end ());		//x - y排序
    ps.erase (unique (ps.begin (), ps.end ()), ps.end ());	//删除重复点
    int n = ps.size (), k = 0;
    vector<Point> qs (n * 2);
    for (int i=0; i<n; ++i) {
        while (k > 1 && cross (qs[k-1] - qs[k-2], ps[i] - qs[k-1]) <= 0)  k--;
        qs[k++] = ps[i];
    }
    for (int i=n-2, t=k; i>=0; --i)  {
        while (k > t && cross (qs[k-1] - qs[k-2], ps[i] - qs[k-1]) <= 0)  k--;
        qs[k++] = ps[i];
    }
    qs.resize (k-1);
    return qs;
}

struct Circle   {
    Point c;
    double r;
    Circle () {}
    Circle (Point c, double r) : c (c), r (r) {}
    Point point(double a)   {
        return Point (c.x + cos (a) * r, c.y + sin (a) * r);
    }
};
struct Line {
    Point p;
    Vector v;
    double r;
    Line () {}
    Line (const Point &p, const Vector &v) : p (p), v (v) {
        r = polar_angle (v);
    }
    Point point(double a)   {
        return p + v * a;
    }
};
/*
    直线相交求交点，返回交点个数，交点保存在P中
*/
int line_cir_inter(Line L, Circle C, double &t1, double &t2, vector<Point> &P)    {
    double a = L.v.x, b = L.p.x - C.c.x, c = L.v.y, d = L.p.y - C.c.y;
    double e = a * a + c * c, f = 2 * (a * b + c * d), g = b * b + d * d - C.r * C.r;
    double delta = f * f - 4 * e * g;
    if (dcmp (delta) < 0)   return 0;
    if (dcmp (delta) == 0)  {
        t1 = t2 = -f / (2 * e); P.push_back (L.point (t1));
        return -1;
    }
    t1 = (-f - sqrt (delta)) / (2 * e); P.push_back (L.point (t1));
    t2 = (-f + sqrt (delta)) / (2 * e); P.push_back (L.point (t2));
    if (dcmp (t1) < 0 || dcmp (t2) < 0) return 0;
    return 2;
}

/*
    两圆相交求交点，返回交点个数。交点保存在P中
*/
int cir_cir_inter(Circle C1, Circle C2, vector<Point> &P)    {
    double d = length (C1.c - C2.c);
    if (dcmp (d) == 0)  {
        if (dcmp (C1.r - C2.r) == 0)    return -1;      //两圆重叠
        else    return 0;
    }
    if (dcmp (C1.r + C2.r - d) < 0) return 0;
    if (dcmp (fabs (C1.r - C2.r) - d) < 0)  return 0;
    double a = polar_angle (C2.c - C1.c);
    double da = acos ((C1.r * C1.r + d * d - C2.r * C2.r) / (2 * C1.r * d));        //C1C2到C1P1的角？
    Point p1 = C1.point (a - da), p2 = C2.point (a + da);
    P.push_back (p1);
    if (p1 == p2)   return 1;
    else    P.push_back (p2);
    return 2;
}
/*
    过点到圆的切线，返回切线条数，切线保存在V中
*/
int point_cir_tan(Point p, Circle C, Vector *V) {
    Vector u = C.c - p;
    double dis = length (u);
    if (dis < C.r)  return 0;
    else if (dcmp (dis - C.r) == 0) {
        V[0] = rotate (u, PI / 2);  return 1;
    }
    else    {
        double ang = asin (C.r / dis);
        V[0] = rotate (u, -ang);
        V[1] = rotate (u, +ang);
        return 0;
    }
}
/*
    两圆的公切线，返回公切线条数，切线短点保存在a和b中
*/
int cir_cir_tan(Circle A, Circle B, Point *a, Point *b) {
    int cnt = 0;
    if (A.r < B.r)  {
        swap (A, B);    swap (a, b);
    }
    double d = dot (A.c - B.c, A.c - B.c);
    double rsub = A.r - B.r, rsum = A.r + B.r;
    if (dcmp (d - rsub) < 0)   return 0;   //内含
    double base = polar_angle (B.c - A.c);
    if (dcmp (d) == 0 && dcmp (A.r - B.r) == 0) return -1;  //两圆重叠
    if (dcmp (d - rsub) == 0)   {       //内切，一条切线
        a[cnt] = A.point (base);    b[cnt] = B.point (base);    cnt++;
        return 1;
    }
    //有外公切线
    double ang = acos (rsub / d);
    a[cnt] = A.point (base + ang);  b[cnt] = B.point (base + ang);  cnt++;
    a[cnt] = A.point (base - ang);  b[cnt] = B.point (base - ang);  cnt++;
    if (d == rsum)  {
        a[cnt] = A.point (base);    b[cnt] = B.point (base + PI);   cnt++;
    }
    else if (dcmp (d - rsum) > 0)   {       //两条内公切线
        double ang2 = acos (rsum / d);
        a[cnt] = A.point (base + ang2); b[cnt] = B.point (base + ang2 + PI);    cnt++;
        a[cnt] = A.point (base - ang2); b[cnt] = B.point (base - ang2 + PI);    cnt++;
    }
    return cnt;
}

Point p1[22], p2[22], p3;

int main(void)    {
    int n;
    while (scanf ("%d", &n) == 1)   {
        if (!n) break;
        for (int i=1; i<=n; ++i)    {
            p1[i] = read_point ();
            p2[i] = Point (p1[i].x, p1[i].y - 1);
        }
        bool flag = false;
        double ans = p1[1].x;
        for (int i=1; i<=n && !flag; ++i)    {
            for (int j=1; j<=n && !flag; ++j)    {
                if (i == j) continue;
                int k;
                for (k=1; k<=n; ++k)    {
                    if (!can_line_seg_inter (p1[i], p2[j], p1[k], p2[k]))   {
                        break;
                    }
                }
                if (k == n + 1)   {
                    flag = true;    break;
                }
                else if (k > max (i, j)) {
                    p3 = line_line_inter (p1[i], p2[j] - p1[i], p1[k-1], p1[k] - p1[k-1]);
                    ans = max (ans, p3.x);
                    p3 = line_line_inter (p1[i], p2[j] - p1[i], p2[k-1], p2[k] - p2[k-1]);
                    ans = max (ans, p3.x);
                }
            }
        }
        if (!flag)  {
            printf ("%.2f\n", ans);
        }
        else    puts ("Through all the pipe.");
    }

   //cout << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n";

    return 0;
}