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\n", name[run (++cas)]);
    }

    return 0;
}

  

 

posted @ 2015-10-22 15:48  Running_Time  阅读(298)  评论(0编辑  收藏  举报