Squares

Time Limit : 7000/3500ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other)

Total Submission(s) : 4   Accepted Submission(s) : 4

Problem 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

 

Source

PKU

 

#include <map>#include <set>#include <list>#include <queue>#include <stack>#include <cmath>#include <ctime>#include <vector>#include <bitset>#include <cstdio>#include <string>#include <numeric>#include <cstring>#include <cstdlib>#include <iostream>#include <algorithm>#include <functional>using namespace std;typedef long long  ll;typedef unsigned long long ull;int dx[4]= {-1,1,0,0};int dy[4]= {0,0,-1,1}; //up down left rightbool inmap(int x,int y,int n,int m){    if(x<1||x>n||y<1||y>m)return false;    return true;}int hashmap(int x,int y,int m){    return (x-1)*m+y;}#define eps 1e-8#define inf 0x7fffffff#define debug puts("BUG")#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define read freopen("in.txt","r",stdin)#define write freopen("out.txt","w",stdout)#define maxn 1111int n;struct Node{    int x,y;} a[maxn];bool operator <(const Node a,const Node b){    if(a.x==b.x)        return a.y<b.y;    else        return a.x<b.x;}int main(){    while(~scanf("%d",&n)&&n)    {        memset(a,0,sizeof(a));        for(int i=0; i<n; i++)            scanf("%d%d",&a[i].x,&a[i].y);        sort(a,a+n);/**    *template <class ForwardIterator, class LessThanComparable>    **bool binary_search(ForwardIterator first, ForwardIterator last,const LessThanComparable& value);    *template <class ForwardIterator, class T, class StrictWeakOrdering>    **bool binary_search(ForwardIterator first, ForwardIterator last, const T& value,StrictWeakOrdering comp);**/        int ans=0;        for(int j=0;j<n;j++)//枚举两个点，再计算出剩下两个点，从原来的集合中进行查找~~用二分查找~~            for(int k=j+1;k<n;k++)/**binary_search()**/        {            Node t;            t.x=a[j].x+a[j].y-a[k].y;//第三个点横坐标：x1+y1-y2            t.y=a[j].y-a[j].x+a[k].x;//        纵坐标：y1-x1+x2            if(!binary_search(a,a+n,t))                continue;            t.x =a[k].x+a[j].y-a[k].y;//第四个点横坐标：x2+y1-y2            t.y =a[k].y-a[j].x+a[k].x;//        纵坐标：y2-x1+x2            if(!binary_search(a,a+n,t))                continue;            ans++;        }        printf("%d\n",ans/2);    }    return 0;}

/**   Hash（转）**/#include<iostream>#include<cmath>#include<algorithm>using namespace std;const int nMax = 1005;const int mMax = 40005;struct data{    int x, y;} node[nMax], tmp;int next[nMax], hash[mMax];    //  学习hash这种存储下标的方法，而不用存详细信息。bool cmp(const data &a, const data &b){    if(a.x == b.x) return a.y < b.y;    return a.x < b.x;}bool search()                  //  寻找是否有点tmp。{    int key = abs(tmp.x + tmp.y);    int i = hash[key];    while(i != -1)    {        if(node[i].x == tmp.x && node[i].y == tmp.y)            return true;        i = next[i];    }    return false;}int main(){    int n, i, j;    while(scanf("%d", &n) && n)    {        memset(hash, -1, sizeof(hash));        for(i = 0; i < n; i ++)            scanf("%d%d", &node[i].x, &node[i].y);        sort(node, node + n, cmp);        for(i = 0; i < n; i ++)        {            int key = abs(node[i].x + node[i].y);            next[i] = hash[key];            hash[key] = i;        }        int ans = 0;        for(i = 0; i < n; i ++)            for(j = i + 1; j < n; j ++)            {                tmp.x = node[i].x + node[i].y - node[j].y;                tmp.y = node[i].y - node[i].x + node[j].x;                if(!search()) continue;                tmp.x = node[j].x + node[i].y - node[j].y;                tmp.y = node[j].y - node[i].x + node[j].x;                if(!search()) continue;                ans ++;            }        printf("%d\n", ans/2);    }    return 0;}