uva 11800 Determine the Shape

　　vjudge上题目链接：Determine the Shape

　　第二道独自 A 出的计算几何水题，题意就是给你四个点，让你判断它是哪种四边形：正方形、矩形、菱形、平行四边形、梯形 or 普通四边形。

　　按照各个四边形的特征层层判断就行，只是这题四个点的相对位置不确定（点与点之间是相邻还是相对并不确定），所以需要枚举各种情况，幸好 4 个点一共只有 3 种不同的情况：abcd、abdc、acbd 画个图就能一目了然，接下来就是编码了，无奈因为一些细节问题我还是调试了还一会 o(╯□╰)o

 #include<cstdio>
 #include<cstring>
 #include<cmath>
 #include<algorithm>
 using namespace std;

 struct Vector {
     double x,y;
     Vector(double x = , double y = ): x(x), y(y) {}
     void readint() {
         int x,y;
         scanf("%d %d",&x,&y);
         this->x = x;
         this->y = y;
     }
     Vector operator + (const Vector &b) const {
         return Vector(x + b.x, y + b.y);
     }
     Vector operator - (const Vector &b) const {
         return Vector(x - b.x, y - b.y);
     }
     Vector operator * (double p) const {
         return Vector(x * p, y * p);
     }
     Vector operator / (double p) const {
         return Vector(x / p, y / p);
     }
 };

 typedef Vector point;
 typedef const point& cpoint;
 typedef const Vector& cvector;

 Vector operator * (double p, const Vector &a) {
     return a * p;
 }

 double dot(cvector a, cvector b) {
     return a.x * b.x + a.y * b.y;
 }

 double length(cvector a) {
     return sqrt(dot(a,a));
 }

 double cross(cvector a, cvector b) {
     return a.x * b.y - a.y *b.x;
 }

 const double eps = 1e-;
 int dcmp(double x) {
     return fabs(x) < eps ? : (x <  ? - : );
 }

 bool operator == (cpoint a, cpoint b) {
     return dcmp(a.x - b.x) ==  && dcmp(a.y - b.y) == ;
 }

 const char s[][] = {"Ordinary Quadrilateral", "Trapezium", "Parallelogram",
                 "Rhombus", "Rectangle", "Square" };

 int judge(cpoint a, cpoint b, cpoint c, cpoint d) {
     if(dcmp(cross(a - b, c - d)) == ) {
         if(dcmp(cross(b - c, a - d)) == ) {
             if(dcmp(length(b - a) - length(d - a)) == ) {
                 if(dcmp(dot(b - a, d - a)) == )    return ;
                 else    return ;
             }
             else if(dcmp(dot(b - a, d - a)) == )   return ;
             else    return ;
         }
         else   return ;
     }
     else if(dcmp(cross(b - c, a - d)) == )    return ;
     else    return ;
 }

 int main() {
     int t,Case = ;
     point a,b,c,d;
     scanf("%d",&t);
     while(t--) {
         a.readint();
         b.readint();
         c.readint();
         d.readint();
         printf("Case %d: ",++Case);
         int ans = ;
         ans = max(ans, judge(a,b,c,d));
         ans = max(ans, judge(a,b,d,c));
         ans = max(ans, judge(a,c,b,d));
         printf("%s\n",s[ans]);
     }
     return ;
 }

　　计算几何，还是很有趣的。。。