POJ 1971 Parallelogram Counting （Hash）

Parallelogram Counting

Time Limit: 5000MS Memory Limit: 65536KTotal Submissions: 5625 Accepted: 1890

Description

There are n distinct points in the plane, given by their integer coordinates. Find the number of parallelograms whose vertices lie on these points. In other words, find the number of 4-element subsets of these points that can be written as {A, B, C, D} such that AB || CD, and BC || AD. No four points are in a straight line.

Input

The first line of the input contains a single integer t (1 <= t <= 10), the number of test cases. It is followed by the input data for each test case. 
The first line of each test case contains an integer n (1 <= n <= 1000). Each of the next n lines, contains 2 space-separated integers x and y (the coordinates of a point) with magnitude (absolute value) of no more than 1000000000. 

Output

Output should contain t lines. 
Line i contains an integer showing the number of the parallelograms as described above for test case i. 

Sample Input

260 02 04 01 13 15 17-2 -18 95 71 14 82 09 8

Sample Output

56

Source

Tehran Sharif 2004 Preliminary

枚举所有对角线，对中点进行Hash即可。

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;#define N 1005#define mod 1000007int x[N],y[N];struct point{    int x,y,cnt;};struct hashtable{    int h[mod],p[mod],size;    point s[mod];    int hash(int x,int y)    {        return ((x*131+y+mod)&0x7FFFFFFF)%mod;    }    void insert(int x,int y)    {        int i,id=hash(x,y);        for(i=h[id]; ~i; i=p[i])            if(s[i].x==x&&s[i].y==y)            {                s[i].cnt++;                return;            }        s[size].cnt=1;        s[size].x=x,s[size].y=y;        p[size]=h[id],h[id]=size++;    }    int find(int x,int y)    {        int i,id=hash(x,y);        for(i=h[id]; ~i; i=p[i])            if(s[i].x==x&&s[i].y==y) return i;        return 0;    }    void clear()    {        size=1;        memset(h,-1,sizeof(h));    }} ht;int main(){    int t,n;    scanf("%d",&t);    while(t--)    {        ht.clear();        scanf("%d",&n);        for(int i=0; i<n; ++i)            scanf("%d%d",x+i,y+i);        for(int i=0; i<n; ++i)            for(int j=i+1; j<n; ++j)                ht.insert(x[i]+x[j],y[i]+y[j]);        int ans=0;        for(int i=0; i<mod; ++i)            for(int j=ht.h[i]; ~j; j=ht.p[j])            {                int tmp=ht.s[j].cnt;                ans+=((tmp-1)*tmp)/2;            }        printf("%d\n",ans);    }    return 0;}