ZOJ2540 Form a Square

Form a Square

题意就是 判断 给你四个点，能否组成一个正方形

要点：

格式很重要， 很重要！！！

数据很小，直接暴力

四个点判断是否为正方形，只需将所有可能的边长度算出来，然后选其中最短的边作为正方形的边长，进行比较，看有没有符合的四条边

 #include<iostream>
 #include<algorithm>
 #include<cmath>
 using namespace std;
 const int N = ;
 typedef struct aa{
     int x, y;
 }AA;
 AA a[N];

 double dis(AA b, AA c)
 {
     double len = sqrt((b.x - c.x) * (b.x - c.x) + (b.y - c.y) * (b.y - c.y));
     return len;
 }
 int main()
 {
     int t, num, cnt = ;
     cin >> t;
     int T = t;
     while(T--)
     {
         for(int i = ; i < ; i++)
         {
             cin >> a[i].x >> a[i].y;
         }
         double ll = ;
         num = ;
         for(int i = ; i < ; i++)
         {
             for(int j = i+; j < ; j++)
             {
                 double dist = dis(a[i], a[j]);
                 if(dist == ll)
                 {
                     num++;
                 }
                 else if(dist < ll)
                 {
                     ll = dist;
                     num = ;
                 }
             }
         }
         if(num == )
         {
             if(cnt != t)
                 cout << "Case " << "" << cnt++ << ":" << endl << "Yes" << endl << endl;
             else
                 cout << "Case " << "" << cnt++ << ":" << endl << "Yes";
         }
         else
         {
             if(cnt != t)
                 cout << "Case " << "" << cnt++ << ":" << endl << "No" << endl << endl;
             else
                 cout << "Case " << "" << cnt++ << ":" << endl << "No";
         }

     }
     return ;
 }