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  • 简单几何(直线求交点) POJ 2074 Line of Sight

    题目传送门

    题意:从一条马路(线段)看对面的房子(线段),问连续的能看到房子全部的最长区间

    分析:自己的思路WA了:先对障碍物根据坐标排序,然后在相邻的障碍物的间隔找到区间,这样还要判断是否被其他障碍物遮挡住(哇

        网上有很好的思路,先对每条线段找到阴影的端点,然后根据坐标排序,求和左端点的距离的最大值,这样省去线段相交的判断。

        trick点应该就是障碍物的位置随意,可能在房子和马路的外面。

    /************************************************
    * Author        :Running_Time
    * Created Time  :2015/11/2 星期一 20:33:38
    * File Name     :POJ_2074.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)    {        //判断直线与线段相交,两点式
        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;
    }
    
    struct Line2 {
        double x1, x2;
        Line2 () {}
        Line2 (double x1, double x2) : x1 (x1), x2 (x2)  {}
        bool operator < (const Line2 &r) const   {
            return x1 < r.x1;
        }
    };
    
    struct Line    {
        Point a, b;
        Line () {}
        Line (Point a, Point b) : a (a), b (b) {}
    };
    
    int main(void)    {
        Line h, p;
        double x1, x2, y;
        while (scanf ("%lf%lf%lf", &x1, &x2, &y) == 3) {
            if (dcmp (x1) == 0 && dcmp (x2) == 0 && dcmp (y) == 0)    break;
            h.a = Point (x1, y);    h.b = Point (x2, y);
            scanf ("%lf%lf%lf", &x1, &x2, &y);
            p.a = Point (x1, y);    p.b = Point (x2, y);
            int n;  scanf ("%d", &n);
            vector<Line2> xs;
            for (int i=0; i<n; ++i) {
                scanf ("%lf%lf%lf", &x1, &x2, &y);
                if (dcmp (y - h.a.y) >= 0 || dcmp (y - p.a.y) <= 0) continue;
                Point p1 = Point (x1, y), p2 = Point (x2, y);
                x1 = line_line_inter (h.b, h.b - p1, p.b, p.b - p.a).x;
                x2 = line_line_inter (h.a, h.a - p2, p.b, p.b - p.a).x;
                if (dcmp (x1 - x2) >= 0)    continue;
                xs.push_back (Line2 (x1, x2));
            }
            xs.push_back (Line2 (p.b.x, p.b.x));
            sort (xs.begin (), xs.end ());
            double left = p.a.x, ans = 0;
            for (int i=0; i<xs.size (); ++i)    {
                x1 = xs[i].x1;  x2 = xs[i].x2;
                if (x1 - left > ans)   ans = x1 - left;
                if (dcmp (x2 - left) > 0)   {
                    left = x2;
                    if (dcmp (left - p.b.x) > 0)   break;
                }
            }
            if (dcmp (ans) == 0)    puts ("No View");
            else    printf ("%.2f
    ", ans);
        }
       
        //cout << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.
    ";
    
        return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/Running-Time/p/4931459.html
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