BestCoder Round #50 (div.1) 1002 Run (HDU OJ 5365) 暴力枚举+正多边形判定

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring> 
 4 #include <algorithm>
 5 #include <cmath>
 6 #define power(x) ((x)*(x))
 7 using namespace std;
 8 const double EPS=1e-8;
 9 const int M = 3e+5;
10 struct node {
11     int x,y;
12 }a[100];
13 int n;
14 int b[100];
15 int dis(int i,int j)    {   //计算2点之间的距离
16     return power(a[i].x-a[j].x)+power(a[i].y-a[j].y);
17 }
18 int main()  {
19     while( ~scanf("%d", &n ) )  {
20         for( int i=0; i<n; i++ )    {
21             scanf("%d %d", &a[i].x, & a[i].y);
22         }
23         int ans = 0;
24         for( int i=0; i<n; i++ )    {
25             for( int j=i+1; j<n; j++ )  {
26                 for( int k=j+1; k<n; k++ )  {
27                     for( int l=k+1; l<n; l++ )  {
28                         b[0] = dis( i, j );
29                         b[1] = dis( i, k );
30                         b[2] = dis( i, l );
31                         b[3] = dis( j, k );
32                         b[4] = dis( j, l );
33                         b[5] = dis( k, l );     //四边形任意2点的距离
34                         sort( b, b+6 );
35                         if( b[0]==b[3] && fabs(b[5]-2*b[0])<EPS && b[4]==b[5] )   {     //判定是否为正方形
36                             ans++;
37                         }
38                     }
39                 }
40             }
41         }
42         printf("%d\n", ans );
43     }
44     return 0;
45 }