POJ

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

2
6
0 0
2 0
4 0
1 1
3 1
5 1
7
-2 -1
8 9
5 7
1 1
4 8
2 0
9 8

Sample Output

5
6

求出最多有几个平行四边形  
两条线段中点相同就是平行四边形  

我一开始是暴力做的  ，正在学哈希  就用哈希写一下  

暴力 

#include<iostream>
#include<cstdio>
#include<algorithm>
#include <cmath>
#include <cstring>
#include <string>
#include<sstream>
#include<set>
#include <cstdlib>
#include<map>
using namespace std;
const int maxn = 1e6 + 10;
struct node {
    double x, y;
    node() {}
    node(double x, double y): x(x), y(y) {}
} a[1010], b[maxn];
int cmp(node a, node b) {
    if (a.x == b.x) return a.y < b.y;
    return a.x < b.x;
}
int main() {
    int t;
    scanf("%d", &t);
    while(t--) {
        int n;
        scanf("%d", &n);
        memset(a, 0, sizeof(a));
        memset(b, 0, sizeof(b));
        for (int i = 0 ; i < n ; i++ )
            scanf("%lf%lf", &a[i].x, &a[i].y);
        int k = 0;
        for (int i = 0 ; i < n - 1 ; i++) {
            for (int j = i + 1 ; j < n ; j++) {
                b[k++] = node((a[i].x + a[j].x) / 2, (a[i].y + a[j].y) / 2);
            }
        }
        sort(b, b + k, cmp);
        int ans = 1, sum = 0;
        for (int i = 0 ; i < k - 1 ; i++) {
            if (b[i].x == b[i + 1].x && b[i].y == b[i + 1].y ) ans++;
            else {
                sum += ans * (ans - 1) / 2;
                ans = 1;
            }
        }
        printf("%d
", sum);
    }
    return 0;
}

哈希

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <string>
#include <sstream>
#include <set>
#include <cstdlib>
#include <map>
using namespace std;
const int maxn = 1e6 + 3;
struct node {
    int x, y, next, sum;
} a[maxn + 10];
int head[maxn], x[1010], y[1010], tot;

int main() {
    int t;
    scanf("%d", &t);
    while(t--) {
        tot = 0;
        memset(head, -1, sizeof(head));
        memset(a, 0, sizeof(a));
        int n;
        scanf("%d", &n);
        int ans = 0;
        for (int i = 0 ; i < n ; i++) {
            int x1, y1;
            scanf("%d%d", &x[i], &y[i]);
            for (int j = 0 ; j < i ; j++) {
                int flag = 1;
                int nx = x[i] + x[j];
                int ny = y[i] + y[j];
                int h = abs(nx + ny) % maxn;
                for (int k = head[h] ; ~k ; k = a[k].next) {
                    if (a[k].x == nx && a[k].y == ny ) {
                        a[k].sum++;
                        ans += a[k].sum;
                        flag = 0;
                        break;
                    }
                }
                if (flag) {
                    a[tot].x = nx;
                    a[tot].y = ny;
                    a[tot].next = head[h];
                    head[h] = tot++;
                }
            }
        }
        printf("%d
", ans);
    }
    return 0;
}