http://poj.org/problem?id=2002
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.
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.
For each test case, print on a line the number of squares one can form from the given stars.
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
1
6
1
题意：给出一些点的集合求能组成正方形的个数。
思路：直接枚举的话$n = 1000\ \ \ O(n^4)$ 会超时。考虑换个思路：已知两个点，由正方形的几何特性
求出另外两个点的坐标，判断是否在点的集合里(哈希，二分，set。。。)都可我用的哈希。
具体先对所有的点按横坐标，纵坐标从小到大排序。统计个数$tot$，最后的答案即为${tot/2}$(正方形的对称性)
那么问题来了，如何已知两个点，求另外两个点呢？
现在给出公式：记$A(x_1,y_1)\ \ \ B(x_2,y_2) \ \ \overrightarrow {AB} =(x_2-x_1,y_2-y_1)$
另外两个点$C(x_3, y_3)\ \ \ D(x_4,y_4)\ \ \ $
$C：\ \ \ x_3 = y_1 - y_2 + x_1\ \ y_3 = x_2 - x_1 + y_1 $
$D：\ \ \ x_4= y_1 - y_2 + x_2\ \ y_4 = x_2 - x_1 + y_2 $
证明其实很简单记$\overrightarrow{X} = (a,b)$逆时针旋转$\beta$度得到$\overrightarrow{Y}(x,y)$
有：$x = acos\beta-bsin\beta\ \ y = asin\beta+bcos\beta$ (由三角函数的几何意义易得)
那么$\overrightarrow {AC} = (x_3-x_1,y_3-y_1)$
$x_3-x_1=(x_2-x_1)cos\beta - (y_2-y_1)sin\beta$
$y_3-y_1=(x_2-x_1)sin\beta + (y_2-y_1)cos\beta$ 其中$\beta = 90^0$
所以$x_3 = y_1-y_2+x_1\ \ y_3=x_2-x_1+y_1$
$\overrightarrow {BD}$同理(注意向量的方向和旋转方向)
原谅我孱弱的语文水平写的太挫了凑合看吧/(ㄒoㄒ)/~~
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<vector>
#include<map>
using std::map;
using std::abs;
using std::sort;
using std::pair;
using std::vector;
using std::multimap;
#define pb(e) push_back(e)
#define sz(c) (int)(c).size()
#define mp(a, b) make_pair(a, b)
#define all(c) (c).begin(), (c).end()
#define iter(c) __typeof((c).begin())
#define cls(arr, val) memset(arr, val, sizeof(arr))
#define cpresent(c, e) (find(all(c), (e)) != (c).end())
#define rep(i, n) for(int i = 0; i < (int)n; i++)
#define tr(c, i) for(iter(c) i = (c).begin(); i != (c).end(); ++i)
const int N = 40007;
const int INF = 0x3f3f3f3f;
struct P {
int x, y;
P() {}
P(int i , int j) :x(i), y(j) {}
inline bool operator<(const P &k) const {
return x == k.x ? y < k.y : x < k.x;
}
}A[N];
struct Hash_Set {
int tot, head[N];
struct edge { int x, y, next; }G[N];
inline void init() {
tot = 0, cls(head, -1);
}
inline void insert(P &k) {
int u = abs(k.x + k.y) % N;
G[tot] = (edge){ k.x, k.y, head[u] }; head[u] = tot++;
}
inline bool find(P k) {
int u = abs(k.x + k.y) % N;
for(int i = head[u]; ~i; i = G[i].next) {
edge &e = G[i];
if(k.x == e.x && k.y == e.y) return true;
}
return false;
}
}hash;
int main() {
#ifdef LOCAL
freopen("in.txt", "r", stdin);
freopen("out.txt", "w+", stdout);
#endif
int n, x1, x2, x3, x4, y1, y2, y3, y4, ans;
while(~scanf("%d", &n), n) {
hash.init();
rep(i, n) {
scanf("%d %d", &A[i].x, &A[i].y);
hash.insert(A[i]);
}
sort(A, A + n);
ans = 0;
for(int i = 0; i < n; i++) {
for(int j = i + 1; j < n; j++) {
x1 = A[i].x, y1 = A[i].y;
x2 = A[j].x, y2 = A[j].y;
x3 = y1 - y2 + x1, y3 = x2 - x1 + y1;
x4 = y1 - y2 + x2, y4 = x2 - x1 + y2;
if(hash.find(P(x3, y3)) && hash.find(P(x4, y4))) ans++;
}
}
printf("%d\n", ans >> 1);
}
return 0;
}
原文：http://www.cnblogs.com/GadyPu/p/4776659.html