• 简单几何(四边形形状) UVA 11800 Determine the Shape


    题目传送门

    题意:给了四个点,判断能构成什么图形,有优先规则

    分析:正方形和矩形按照点积为0和长度判断,菱形和平行四边形按向量相等和长度判断,梯形按照叉积为0判平行。因为四个点是任意给出的,首先要进行凸包排序,可能会有三点共线的情况。

    /************************************************
    * Author        :Running_Time
    * Created Time  :2015/10/22 星期四 13:27:33
    * File Name     :UVA_11800.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;
    struct Point    {       //点的定义
        double x, y;
        Point (double x=0, double y=0) : x (x), y (y) {}
    };
    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 polar_angle(Vector A)  {     //向量极角
        return atan2 (A.y, A.x);
    }
    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;
    }
    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;
    }
    Vector operator + (Vector A, Vector B)  {       //向量加法
        return Vector (A.x + B.x, A.y + B.y);
    }
    Vector operator - (Vector A, Vector B)  {       //向量减法
        return Vector (A.x - B.x, A.y - B.y);
    }
    Vector operator * (Vector A, double p)  {       //向量乘以标量
        return Vector (A.x * p, A.y * p);
    }
    Vector operator / (Vector A, double p)  {       //向量除以标量
        return Vector (A.x / p, A.y / p);
    }
    bool operator < (const Point &a, const Point &b)    {       //点的坐标排序
        return a.x < b.x || (a.x == b.x && a.y < b.y);
    }
    bool operator == (const Point &a, const Point &b)   {       //判断同一个点
        return dcmp (a.x - b.x) == 0 && dcmp (a.y - b.y) == 0;
    }
    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));
    }
    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;
    }
    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 point_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 dis_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 dis_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_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 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_poly(Point *p, int n)   {       //多边形面积
        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,输入点集会被修改
    */
    vector<Point> convex_hull(vector<Point> &P) {
        sort (P.begin (), P.end ());
        P.erase (unique (P.begin (), P.end ()), P.end ());      //预处理,删除重复点
        int n = P.size (), m = 0;
        vector<Point> ret (n + 1);
        for (int i=0; i<n; ++i) {
            while (m > 1 && cross (ret[m-1]-ret[m-2], P[i]-ret[m-2]) < 0)   m--;
            ret[m++] = P[i];
        }
        int k = m;
        for (int i=n-2; i>=0; --i)  {
            while (m > k && cross (ret[m-1]-ret[m-2], P[i]-ret[m-2]) < 0)   m--;
            ret[m++] = P[i];
        }
        if (n > 1)  m--;
        ret.resize (m);
        return ret;
    }
    
    char name[6][30] = {
        "Square", "Rectangle", "Rhombus", "Parallelogram", "Trapezium", "Ordinary Quadrilateral"
    };
    
    int run(int cas)    {
        printf ("Case %d: ", cas);
        vector<Point> P;
        for (int i=0; i<4; ++i) {
            P.push_back (read_point ());
        }
        vector<Point> Ps = convex_hull (P);
        if (Ps.size () != 4)    return 5;
        Point &a = Ps[0], &b = Ps[1], &c = Ps[2], &d = Ps[3];
        /*  a d
            b c    */
        Vector ba = b - a, da = d - a, cd = c - d, cb = c - b;
        double lba = length (ba), lcd = length (cd), lda = length (da), lcb = length (cb);
        if (dot (ba, da) == 0 && dot (da, cd) == 0 && dot (cb, cd) == 0)    {
            if (lba == lcb) return 0;
            else    return 1;
        }
        if (ba == cd && da == cb)   {
            if (lba == lda) return 2;
            else    return 3;
        }
        if (cross (ba, cd) == 0 || cross (da, cb) == 0) return 4;
        else    return 5;
    }
    
    
    
    int main(void)    {
        Point a, b, c, d;
        int T, cas = 0; scanf ("%d", &T);
        while (T--) {
            printf ("%s
    ", name[run (++cas)]);
        }
    
        return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/Running-Time/p/4901271.html
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