poj_2002 Squares（哈希）

Squares

Time Limit: 3500MS Memory Limit: 65536KTotal Submissions: 19231 Accepted: 7429

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

题目要求找直角坐标系中所给点一共能组成多少个正方形。

在纸上模拟一下，比较容易发现已知两点(x1,y1),(x2,y2)能求出正方形中的另外两点(x3,y3),(x4,y4)

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

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

然后就可以用哈希表保存所有的点，然后枚举两点去找另外两点。

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <stack>#include <bitset>#include <queue>#include <set>#include <map>#include <string>#include <algorithm>#define FOP freopen("data.txt","r",stdin)#define inf 0x3f3f3f3f#define maxn 1000010#define mod 1000007#define PI acos(-1.0)#define LL long longusing namespace std;struct Node{    int x, y;    int next;}node[maxn];int n;int cot;int hashTable[maxn];void initHash(){    cot = 0;    memset(hashTable, -1, sizeof(hashTable));}int getHash(int x, int y){    return ((x * x) % mod + (y * y) % mod) % mod;}void insertHash(int x, int y){    int key = getHash(x, y);    node[cot].x = x, node[cot].y = y;    node[cot].next = hashTable[key];    hashTable[key] = cot++;}bool searchHash(int x, int y){    int key = getHash(x, y);    int next = hashTable[key];    while(next != -1)    {        if(x == node[next].x && y == node[next].y) return true;        next = node[next].next;    }    return false;}struct Point{    int x, y;}p[1010];int main(){    while(scanf("%d", &n) && n)    {        int ans = 0;        initHash();        for(int i = 1; i <= n; i++) scanf("%d%d", &p[i].x, &p[i].y), insertHash(p[i].x, p[i].y);        for(int i = 1; i <= n; i++)        {            for(int j = i+1; j <= n; j++)            {                int x1 = p[i].x + (p[j].y - p[i].y), y1 = p[i].y - (p[j].x - p[i].x);                int x2 = p[j].x + (p[j].y - p[i].y), y2 = p[j].y - (p[j].x - p[i].x);                int x3 = p[i].x - (p[j].y - p[i].y), y3 = p[i].y + (p[j].x - p[i].x);                int x4 = p[j].x - (p[j].y - p[i].y), y4 = p[j].y + (p[j].x - p[i].x);                if((searchHash(x1, y1) && searchHash(x2, y2))) ans++;                if((searchHash(x3, y3) && searchHash(x4, y4))) ans++;            }        }        printf("%d\n", ans / 4);    }    return 0;}