POJ 2002 Squares [hash]

Squares

Time Limit: 3500MS		Memory Limit: 65536K
Total Submissions: 16631		Accepted: 6328

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

4
1 0
0 1
1 1
0 0
9
0 0
1 0
2 0
0 2
1 2
2 2
0 1
1 1
2 1
4
-2 5
3 7
0 0
5 2
0

Sample Output

1
6
1

Source

Rocky Mountain 2004

题解：枚举两个点，算出另外两点坐标，看是否在给的点集里。

具体实现用hash，hash碰撞了就放在链表中，然后在链表里查找~（天猫所说的  pascal拉链哈希，，，

就写了个哈希函数然后对函数值指针拉链出来哈希）

）

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<map>
#define ll long long
#define N 1005
#define mod 2007

using namespace std;

int n;
int x[2*N];
int y[2*N];
//map<pair<int,int>,int>mp;
int ans;
int head[2*N];
int next[2*N];
int m;

void insert(int i)
{
    int key=(x[i]*x[i]+y[i]*y[i])%mod;
    next[m]=head[key];
    x[m]=x[i];
    y[m]=y[i];
    head[key]=m++;
}

void ini()
{
    ans=0;
    m=N;
    memset(head,-1,sizeof(head));
    //mp.clear();
    int i;
    for(i=1;i<=n;i++){
        scanf("%d%d",&x[i],&y[i]);
        insert(i);
        //mp[ make_pair(x[i],y[i]) ]=i;
    }
}

int find(int xx,int yy)
{
    int key=(xx*xx+yy*yy)%mod;
    int i;
    for(i=head[key];i!=-1;i=next[i]){
        if(x[i]==xx && y[i]==yy){
            return 1;
        }
    }
    return 0;
}

int ok(int i,int j)
{
    int mx,my;
    int x3,x4,y3,y4;
    int he,cha;
    mx=x[i]+x[j];
    my=y[i]+y[j];
    he=mx+my;cha=my-mx;
    if(he%2!=0 || cha%2!=0) return 0;
    he/=2;cha/=2;
    x3=he-y[i];
   // x3=mx+my-y[i];
    y3=cha+x[i];
 //   y3=my-(mx-x[i]);
    if(find(x3,y3)==0) return 0;
    x4=y[i]-cha;
   // x4=mx-(my-y[i]);
   // y4=my+(mx-x[i]);
    y4=he-x[i];
    if(find(x4,y4)==0) return 0;

    return 1;
}

void solve()
{
    int i,j;
    for(i=1;i<n;i++){
        for(j=i+1;j<=n;j++){
            if(ok(i,j)==1){
                ans++;
            }
        }
    }
}

void out()
{
    ans/=2;
    printf("%d
",ans);
}

int main()
{
    //freopen("data.in","r",stdin);
   // scanf("%d",&T);
    //while(T--){
    while(scanf("%d",&n)!=EOF){
        if(n==0) break;
        ini();
        solve();
        out();
    }
    return 0;
}