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  • UVA 11800 Determine the Shape --凸包第一题

    题意: 给四个点,判断四边形的形状。可能是正方形,矩形,菱形,平行四边形,梯形或普通四边形。

    解法: 开始还在纠结怎么将四个点按序排好,如果直接处理的话,有点麻烦,原来凸包就可搞,直接求个凸包,然后点就自动按逆时针排好了,然后就判断就可以了,判断依据题目下面有,主要是用到点积和叉积,判断垂直用点积,判断平行用叉积。

    代码:

    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <cstdlib>
    #include <cmath>
    #include <algorithm>
    #define eps 1e-8
    using namespace std;
    
    struct Point{
        double x,y;
        Point(double x=0, double y=0):x(x),y(y) {}
        void input() { scanf("%lf%lf",&x,&y); }
    };
    typedef Point Vector;
    struct Circle{
        Point c;
        double r;
        Circle(){}
        Circle(Point c,double r):c(c),r(r) {}
        Point point(double a) { return Point(c.x + cos(a)*r, c.y + sin(a)*r); }
        void input() { scanf("%lf%lf%lf",&c.x,&c.y,&r); }
    };
    struct Line{
        Point p;
        Vector v;
        double ang;
        Line(){}
        Line(Point p, Vector v):p(p),v(v) { ang = atan2(v.y,v.x); }
        Point point(double t) { return Point(p.x + t*v.x, p.y + t*v.y); }
        bool operator < (const Line &L)const { return ang < L.ang; }
    };
    int dcmp(double x) {
        if(x < -eps) return -1;
        if(x > eps) return 1;
        return 0;
    }
    template <class T> T sqr(T x) { return x * x;}
    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); }
    bool operator < (const Point& a, const Point& b) { return a.x < b.x || (a.x == b.x && a.y < b.y); }
    bool operator >= (const Point& a, const Point& b) { return a.x >= b.x && a.y >= b.y; }
    bool operator <= (const Point& a, const Point& b) { return a.x <= b.x && a.y <= b.y; }
    bool operator == (const Point& a, const Point& b) { return dcmp(a.x-b.x) == 0 && dcmp(a.y-b.y) == 0; }
    double Dot(Vector A, Vector B) { return A.x*B.x + A.y*B.y; }
    double Length(Vector A) { return sqrt(Dot(A, A)); }
    double Angle(Vector A, Vector B) { return acos(Dot(A, B) / Length(A) / Length(B)); }
    double Cross(Vector A, Vector B) { return A.x*B.y - A.y*B.x; }
    Vector VectorUnit(Vector x){ return x / Length(x);}
    Vector Normal(Vector x) { return Point(-x.y, x.x) / Length(x);}
    double angle(Vector v) { return atan2(v.y, v.x); }
    int ConvexHull(Point* p, int n, Point* ch)
    {
        sort(p,p+n);
        int m = 0;
        for(int i=0;i<n;i++) {
            while(m > 1 && Cross(ch[m-1]-ch[m-2], p[i]-ch[m-2]) <= 0) m--;
            ch[m++] = p[i];
        }
        int k = m;
        for(int i=n-2;i>=0;i--) {
            while(m > k && Cross(ch[m-1]-ch[m-2], p[i]-ch[m-2]) <= 0) m--;
            ch[m++] = p[i];
        }
        if(n > 1) m--;
        return m;
    }
    
    //data segment
    Point p[5],ch[5];
    Point A,B,C,D;
    //data ends
    
    int main()
    {
        int t,n,i,cs = 1;
        scanf("%d",&t);
        while(t--)
        {
            for(i=0;i<4;i++) p[i].input();
            printf("Case %d: ",cs++);
            int m = ConvexHull(p,4,ch);
            if(m < 4) { puts("Ordinary Quadrilateral"); continue; }
            A = ch[0], B = ch[1], C = ch[2], D = ch[3];
    
            if(dcmp(Dot(B-A,D-A)) == 0 && dcmp(Dot(B-A,C-B)) == 0 && dcmp(Dot(C-B,C-D)) == 0 && dcmp(Dot(D-C,D-A)) == 0
            && dcmp(Length(B-A)-Length(C-B)) == 0 && dcmp(Length(C-B)-Length(D-C)) == 0 && dcmp(Length(C-D)-Length(A-D)) == 0)
                puts("Square");
            else if(dcmp(Dot(B-A,D-A)) == 0 && dcmp(Dot(B-A,C-B)) == 0 && dcmp(Dot(C-B,C-D)) == 0 && dcmp(Dot(D-C,D-A)) == 0
                 && dcmp(Length(A-D)-Length(C-B)) == 0 && dcmp(Length(A-B)-Length(C-D)) == 0)
                puts("Rectangle");
            else if(dcmp(Length(B-A)-Length(C-B)) == 0 && dcmp(Length(C-B)-Length(D-C)) == 0 && dcmp(Length(C-D)-Length(A-D)) == 0)
                puts("Rhombus");
            else if(dcmp(Length(A-D)-Length(B-C)) == 0 && dcmp(Length(A-B)-Length(C-D)) == 0)
                puts("Parallelogram");
            else if(dcmp(Cross(B-C,D-A)) == 0 || dcmp(Cross(B-A,D-C)) == 0)
                puts("Trapezium");
            else
                puts("Ordinary Quadrilateral");
        }
        return 0;
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/whatbeg/p/4158726.html
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