学步园

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;
}