poj2002 Squares--哈希表

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

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

Sample Output

161

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 1000005using 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;}