题意：给定四个点，判断形状
思路：先求个凸包，就能把四个点排序，然后就是利用几何去判断，利用点积判垂直，利用叉积判平行
还有这题有个坑啊，明明说好是没有点共线的，其实是有的，所以求凸包如果不是4个点，直接输出不规则四边形即可
代码：
#include <cstdio>#include <cstring>#include<cstdio>#include<cmath>#include<algorithm>using namespace std;const int MAXN = 4;struct Point {int x, y;Point() {}Point(double x, double y) {this->x = x;this->y = y;}void read() {scanf("%d%d", &x, &y);}} list[MAXN], p[MAXN];typedef Point Vector;Vector operator + (Vector A, Vector B) {return Vector(A.x + B.x, A.y + B.y);}Vector operator – (Vector A, Vector B) {return Vector(A.x – B.x, A.y – B.y);}Vector operator * (Vector A, double p) {return Vector(A.x * p, A.y * p);}Vector operator / (Vector A, double p) {return Vector(A.x / p, A.y / p);}int Dot(Vector A, Vector B) {return A.x * B.x + A.y * B.y;} //点积int Cross(Vector A, Vector B) {return A.x * B.y – A.y * B.x;} //叉积int stack[MAXN], top;int xmult(Point p0, Point p1, Point p2) {return (p1.x – p0.x) * (p2.y – p0.y) – (p1.y – p0.y) * (p2.x – p0.x);}double dist(Point p1,Point p2) {return sqrt((double)(p2.x-p1.x)*(p2.x-p1.x)+(p2.y-p1.y)*(p2.y-p1.y));}int dist2(Point p1, Point p2) {return (p2.x – p1.x) * (p2.x – p1.x) + (p2.y – p1.y) * (p2.y – p1.y);}//极角排序函数 ，， 角度相同则距离小的在前面bool cmp(Point p1,Point p2) {int tmp= xmult(list[0], p1, p2);if(tmp > 0) return true;else if(tmp == 0 && dist(list[0], p1) < dist(list[0], p2)) return true;else return false;}bool LineParallel(Vector v, Vector w) {return Cross(v, w) == 0;}bool LineVertical(Vector v, Vector w) {return Dot(v, w) == 0;}//输入并把最左下方的点放在list[0]并且进行极角排序void init(int n) {int i, k;Point p0;scanf("%d%d", &list[0].x, &list[0].y);p0.x = list[0].x;p0.y = list[0].y;k = 0;for(i = 1; i < n; i++) {scanf("%d%d", &list[i].x, &list[i].y);if((p0.y > list[i].y) || ((p0.y == list[i].y) && (p0.x > list[i].x))) {p0.x = list[i].x;p0.y = list[i].y;k = i;}}list[k] = list[0];list[0] = p0;sort(list + 1, list + n, cmp);}void graham(int n) {int i;if(n == 1) {top = 0; stack[0] = 0;}if(n == 2) {top = 1;stack[0] = 0;stack[1] = 1;}if(n > 2) {for(i = 0; i <= 1; i++) stack[i] = i;top = 1;for(i = 2; i < n; i++) {while(top > 0 && xmult(list[stack[top – 1]], list[stack[top]], list[i]) <= 0) top–;top++;stack[top]=i;}}for (int i = 0; i <= top; i++)p[i] = list[stack[i]];}int t;void solve() {if (LineParallel(p[1] – p[0], p[2] – p[3]) && LineParallel(p[2] – p[1], p[3] – p[0])) {if (LineVertical(p[1] – p[0], p[2] – p[1]) && LineVertical(p[2] – p[1], p[3] – p[2]) && LineVertical(p[0] – p[3], p[3] – p[2])) {if (dist2(p[1], p[0]) == dist2(p[2], p[1])) printf("Square\n");else printf("Rectangle\n");} else {if (dist2(p[1], p[0]) == dist2(p[2], p[1])) printf("Rhombus\n");else printf("Parallelogram\n");}} else if (LineParallel(p[1] – p[0], p[2] – p[3]) || LineParallel(p[2] – p[1], p[3] – p[0])) printf("Trapezium\n");else printf("Ordinary Quadrilateral\n");}int main() {int cas = 0;scanf("%d", &t);while (t–) {init(4);graham(4);printf("Case %d: ", ++cas);if (top < 3) {printf("Ordinary Quadrilateral\n");continue;}for (int i = 0; i < 4; i++)p[i] = list[stack[i]];solve();}return 0;}
未经一番寒彻骨，焉得梅花扑鼻香
