Squares POJ2002 哈希的应用

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

41 00 11 10 090 01 02 00 21 22 20 11 12 14-2 53 70 05 20

Sample Output

161

题意：

有一堆平面散点集，任取四个点，求能组成正方形的不同组合方式有多少。

相同的四个点，不同顺序构成的正方形视为同一正方形。

思路：

暴力枚举四个点肯定超时， pass

先枚举两个点， 然后再（乱搞）计算出另外两个点的坐标， 在哈希表中查找是否存在这两个坐标（存在就代表可以画出一个正方形）。因为每条边都计算了一次， 所以最后结果除以4。 

大家可以先推推两外两个点的坐标， 下面给出结果：

已知： (x1,y1)  (x2,y2)

则：   

x3 = x1 + (y1-y2)   y3 = y1 - (x1-x2)   x4 = x2 + (y1-y2)   y4 = y2 -(x1-x2)

或

x3 = x1 - (y1-y2)   y3 = y1 + (x1-x2)  x4 = x2 - (y1-y2)   y4 = y2 + (x1-x2)

其实还可以 已知一条对角线上两个点的坐标， 计算出另外两个点的坐标。 结果就不写了， 大家可以去推推

哈希表用数组模拟链表。 具体看代码

AC code： 1125ms

#include <iostream>#include <cstring>#include <algorithm>using namespace std;struct  Node{    int x, y, next;}Hash[1200];int cur, ans, hashTable[10011];void init() {    cur = ans = 0;    memset(hashTable, -1, sizeof(hashTable));}void insert(int x, int y) {    int number = (x*x + y*y) % 10011;    Hash[cur].x = x;    Hash[cur].y = y;    Hash[cur].next = hashTable[number];    hashTable[number] = cur;    cur++;}bool search(int x, int y) {    int number = (x*x + y*y) % 10011;    int next = hashTable[number];    while (next != -1) {        if (Hash[next].x == x && Hash[next].y == y) return true;        next = Hash[next].next;    }    return false;}int main(void) {    int n;    while (cin >> n , n) {        init();        int x[1200], y[1200];        for (int i = 0; i < n; i++) {            cin >> x[i] >> y[i];            insert(x[i], y[i]);        }                for (int i = 0; i < n; i++) {            for (int j = i+1; j < n; j++) {                int x1 = x[i] - y[i] + y[j], y1 = y[i] + x[i] - x[j];                int x2 = x[j] - y[i] + y[j], y2 = y[j] + x[i] - x[j];                if (search(x1, y1) && search(x2, y2)) ans++;                int x3 = x[i] + y[i] - y[j], y3 = y[i] - x[i] + x[j];                int x4 = x[j] + y[i] - y[j], y4 = y[j] - x[i] + x[j];                if (search(x3, y3) && search(x4, y4)) ans++;            }        }                cout << ans/4 << endl;    }    return 0;}