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[*N];
 int y[*N];
 //map<pair<int,int>,int>mp;
 int ans;
 int head[*N];
 int next[*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=;
     m=N;
     memset(head,-,sizeof(head));
     //mp.clear();
     int i;
     for(i=;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!=-;i=next[i]){
         if(x[i]==xx && y[i]==yy){
             return ;
         }
     }
     return ;
 }

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

     return ;
 }

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

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

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