原题链接：​​http://poj.org/problem?id=2002​​


一：原题内容

Description


A square is a 4-sided polygon whose sides have equal length and adjacent sides form 90-degree angles. It is also a polygon such that rotating about its centre by 90 degrees gives the same polygon. It is not the only polygon with the latter property, however, as a regular octagon also has this property.

So we all know what a square looks like, but can we find all possible squares that can be formed from a set of stars in a night sky? To make the problem easier, we will assume that the night sky is a 2-dimensional plane, and each star is specified by its x and y coordinates.


Input


The input consists of a number of test cases. Each test case starts with the integer n (1 <= n <= 1000) indicating the number of points to follow. Each of the next n lines specify the x and y coordinates (two integers) of each point. You may assume that the points are distinct and the magnitudes of the coordinates are less than 20000. The input is terminated when n = 0.


Output


For each test case, print on a line the number of squares one can form from the given stars.


Sample Input


4
1 0
0 1
1 1
0 0
9
0 0
1 0
2 0
0 2
1 2
2 2
0 1
1 1
2 1
4
-2 5
3 7
0 0
5 2
0


Sample Output


1 6 1


二：分析理解

在坐标系，给出n个点坐标，求能组成多少个正方形？数据保证没有相同的点坐标。

枚举两个点，求出另外正方形两点，看是否存在。

三：AC代码

#include<iostream>    
#include<string.h>  
#include<algorithm>  
#include<cmath>

#define N 1000005
using namespace std;

struct Node
{
  int x;
  int y;
};


bool cmp(Node node1, Node node2)
{
  if (node1.x == node2.x)
    return node1.y < node2.y;
  return node1.x < node2.x;
}

Node node[N];
int head[N];
int next1[N];
int n;
int m;

void Insert(int i)
{
  int key = (node[i].x*node[i].x + node[i].y*node[i].y) % N;
  next1[m] = head[key];//这里之所以有m，因为下面的sort操作，改变了原始的node数组，所以下面注释的代码不能用
  node[m].x = node[i].x;
  node[m].y = node[i].y;
  head[key] = m++;

  //next1[i]=head[key];
  //head[key]=i;
}

int Find(int x,int y)
{
  int key = (x*x + y*y) % N;
  for (int i = head[key]; i != -1; i = next1[i])
    if (node[i].x == x&&node[i].y == y)
      return i;
  return -1;
}

int main()
{
  while (scanf("%d", &n) && n)
  {
    memset(head, -1, sizeof(head));
    memset(next1, -1, sizeof(next1));
    m = 1005;
    for (int i = 0; i < n; i++)
    {
      scanf("%d%d", &node[i].x, &node[i].y);
      Insert(i);
    }

    sort(node, node + n, cmp);
    int ans = 0;
    for (int i = 0; i <= n - 2; i++)
    {
      for (int j = i + 1; j <= n - 1; j++)
      {
        int x1, y1;
        int x2, y2;
        int width1 = node[i].y - node[j].y;
        int width2 = node[j].x - node[i].x;

        if (width1 > 0)
        {
          x1 = node[i].x - width1;
          y1 = node[i].y - width2;
          if (Find(x1, y1) == -1)//没找到
            continue;

          x2 = node[j].x - width1;
          y2 = node[j].y - width2;
          if (Find(x2, y2) == -1)
            continue;

          ans++;
        }
        else
        {
          width1 = -width1;
          x1 = node[j].x - width1;
          y1 = node[j].y + width2;
          if (Find(x1, y1) == -1)
            continue;

          x2 = node[i].x - width1;
          y2 = node[i].y + width2;
          if (Find(x2, y2) == -1)
            continue;

          ans++;
        }
      }
    }

    printf("%d\n", ans / 2);
  }

  return 0;
}