zoukankan      html  css  js  c++  java
  • 简单几何(凸包) POJ 1696 Space Ant

    题目传送门

    题意:一个蚂蚁一直往左边走,问最多能走多少步,且输出路径

    分析:就是凸包的变形题,凸包性质,所有点都能走。从左下角开始走,不停排序。有点纠结,自己的凸包不能AC。待理解透凸包再来写。。 好像只能用卷包裹法来写,就是从一个起点出发,每次相对于起点用叉积排序,选择最外侧的点,更新起点。

    /************************************************
    * Author        :Running_Time
    * Created Time  :2015/10/27 星期二 14:20:36
    * File Name     :POJ_1696.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;
        int id;
        Point (double x=0, double y=0, int id = 0) : x (x), y (y), id (id) {}
        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)  {       //向量乘以标量
            return Point (x * p, y * p);
        }
        Point operator / (double p)  {       //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
        int id;
        double x, y;
        scanf ("%d%lf%lf", &id, &x, &y);
        return Point (x, y, id);
    }
    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_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 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;
    }
    int pos;
    bool vis[55];
    Point p[55];
    bool cmp(Point a, Point b)  {
        double t = cross (a - p[pos], b - p[pos]);
        if (dcmp (t) == 0)  {
            return length (a - p[pos]) < length (b - p[pos]);
        }
        else if (dcmp (t) < 0)  return false;
        else    return true;
    }
    
    /*
        点集凸包
    */
    vector<int> convex_hull(vector<Point> P) {
        sort (P.begin (), P.end ());
        int n = P.size (), k = 0;
        vector<Point> ret (n * 2);
        vector<int> id (n * 2);
        for (int i=0; i<n; ++i) {
            while (k > 1 && cross (ret[k-1] - ret[k-2], P[i] - ret[k-1]) <= 0)  k--;
            id[k] = P[i].id;
            ret[k++] = P[i];
        }
        for (int i=n-2, t=k; i>=0; --i)  {
            while (k > t && cross (ret[k-1] - ret[k-2], P[i] - ret[k-1]) <= 0)  k--;
            id[k] = P[i].id;
            ret[k++] = P[i];
        }
        ret.resize (k-1);   id.resize (k - 1);
        return id;
    }
    
    int main(void)    {
        int T;  scanf ("%d", &T);
        while (T--) {
            int n;  scanf ("%d", &n);
            for (int i=0; i<n; ++i) {
                p[i] = read_point ();
                if (p[0].y > p[i].y || (p[0].y == p[i].y && p[0].x > p[i].x))   swap (p[0], p[i]);
            }
            pos = 0;
            for (int i=1; i<n; ++i) {
                sort (p+i, p+n, cmp);
                pos++;
            }
            printf ("%d", n);
            for (int i=0; i<n; ++i) {
                printf (" %d", p[i].id);
            }
            puts ("");
        }
    
       //cout << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.
    ";
    
        return 0;
    }
    

      

    编译人生,运行世界!
  • 相关阅读:
    图片的使用
    对话框
    窗体
    浏览器与android移动端视频互播技术实现
    Arcengine实现创建网络数据集札记(三)
    Arcengine实现创建网络数据集札记(二)
    Arcengine实现创建网络数据集札记(一)
    2019年年初iOS招人心得笔记(附面试题)
    2019年,200道面试题打造最受企业欢迎的iOS程序猿!
    BAT面试总结——iOS开发高级工程师
  • 原文地址:https://www.cnblogs.com/Running-Time/p/4914922.html
Copyright © 2011-2022 走看看