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; }
简单几何(四边形形状) UVA 11800 Determine the Shape
原文:http://www.cnblogs.com/Running-Time/p/4901271.html