poj 2002 Squares

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

哈希表做法，程序略渣#include <iostream>#include <cstring>#include <cstdio>#include <cstring>#include <algorithm>#include <queue>#include <stack>#include <cmath>#define LL long long#define INF 0x3f3f3f3f#define Mod 1000007using namespace std;struct node{    int x,y;    int next;} ls[1010];int top;int Hash[Mod];int x[1010],y[1010];void Add(int x,int y){    int a = (x * x + y * y) % Mod;    ls[top].x = x;    ls[top].y = y;    ls[top].next = Hash[a];    Hash[a] = top++;}bool Search(int x,int y){    int a = (x * x + y * y) % Mod;    int next = Hash[a];    while(next != -1 )    {        if(ls[next].x == x && ls[next].y == y)            return true;        next = ls[next].next;    }    return false;}int main(){   // freopen("in.txt","r",stdin);    int n;    while(cin>>n && n)    {        top = 0;        memset(Hash,-1,sizeof(Hash));        for(int i=0; i<n; i++)        {            cin>>x[i]>>y[i];            Add(x[i],y[i]);        }        LL ant = 0;        for(int i=0; i<n-1; i++)        {            for(int j=i+1; j<n; j++)            {                int xx = x[i] - x[j];                int yy = y[i] - y[j];                int x3 = x[i] + yy;                int y3 = y[i] - xx;                int x4 = x[j] + yy;                int y4 = y[j] - xx;                if(Search(x3,y3) && Search(x4,y4))                    ant++;                x3 = x[i] - yy;                y3 = y[i] + xx;                x4 = x[j] - yy;                y4 = y[j] + xx;                if(Search(x3,y3) && Search(x4,y4))                    ant++;            }        }        cout<<ant/4<<endl;    }    return 0;}//二分查找<pre name="code" class="html">#include <iostream>#include <cstring>#include <cstdio>#include <cstring>#include <algorithm>#include <queue>#include <stack>#include <cmath>#define LL long long#define INF 0x3f3f3f3f#define Mod 1000007using namespace std;struct node{    int x,y;} ls[1010];bool cmp(node a,node b){    if(a.x == b.x)        return a.y < b.y;    return a.x < b.x;}bool Search(int low,int high,int x,int y){    int i = low,j = high;    while(i <= j)    {        int mid = (i + j) / 2;        if(ls[mid].x == x && ls[mid].y == y)            return true;        if(ls[mid].x < x)            i = mid + 1;        else if(ls[mid].x > x)            j = mid - 1;        else if(ls[mid].x == x)        {            if(ls[mid].y < y)                i = mid + 1;            else if(ls[mid].y > y)                j = mid - 1;        }    }    return false;}int main(){    std::ios::sync_with_stdio(false);    int n;    while(cin>>n && n)    {        for(int i=0; i<n; i++)        {            cin>>ls[i].x>>ls[i].y;        }        sort(ls,ls+n,cmp);        int ant = 0;        for(int i=0; i<n-1; i++)        {            for(int j=i+1; j<n; j++)            {                int xx = ls[i].x - ls[j].x;                int yy = ls[i].y - ls[j].y;                int x3 = ls[i].x - yy;                int y3 = ls[i].y + xx;                int x4 = ls[j].x - yy;                int y4 = ls[j].y + xx;                if(Search(0,n-1,x3,y3) && Search(0,n-1,x4,y4))                    ant++;                x3 = ls[i].x + yy;                y3 = ls[i].y - xx;                x4 = ls[j].x + yy;                y4 = ls[j].y - xx;                if(Search(0,n-1,x3,y3) && Search(0,n-1,x4,y4))                    ant++;            }        }        cout<<ant/4<<endl;    }    return 0;}