poj 2202 Squares(求正方形个数)

Squares
Time Limit: 3500MS Memory Limit: 65536K
Total Submissions: 18162 Accepted: 6969
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

#include <cstdio>#include <cstring>#include<algorithm>#include <cmath>#include <vector>using namespace std;#define N 1000int n;struct node {int x;int y;}point[N];bool cmp(node a, node b){return (a.x < b.x || (a.x == b.x && a.y < b.y));}int bsearch(int x, int y){intmid, l = 0, r = n - 1;while (l <= r) {mid = (l + r) / 2;if (point[mid].x == x && point[mid].y == y) return 1;if (point[mid].x > x || (point[mid].x == x && point[mid].y > y)) {r = mid - 1;}else {l = mid + 1;}}return 0;}int main(){int x, y, i, j, count;while (scanf("%d", &n)&&n){count = 0;for (i = 0; i < n; i++)scanf("%d %d", &point[i].x, &point[i].y);sort(point, point + n, cmp);for (i = 0; i < n; i++) {for (j = (i + 1); j < n; j++) {x = point[i].y - point[j].y + point[i].x;y = point[j].x - point[i].x + point[i].y;if (bsearch(x, y) == 0) {continue;}x = point[i].y - point[j].y + point[j].x;y = point[j].x - point[i].x + point[j].y;if (bsearch(x, y)) {count++;}}}printf("%d\n", count / 2);}return 0;}