Trapezium

Problem Statement

There are $N$ points on a two-dimensional plane, with the $i$-th point at coordinates $(X_i, Y_i)$. It is guaranteed that no two points are at the same position, and no three points are collinear.

Among the combinations of four points from these points, how many combinations can form a trapezoid as a polygon with those four points as vertices?

Constraints

$4 \leq N \leq 2,000$
$0 \leq X_i, Y_i \leq 10^7$ ($1 \leq i \leq N$)
No two points are at the same location.
No three points are collinear.
All input values are integers.

Solving

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void solve() 
{
    int n;
    cin >> n;
    struct Point 
    {
        int x,y;
    };
    vector<Point> p(n+1);
    for(int i=1;i<=n;i++) 
    {
        cin >> p[i].x >> p[i].y;
    }
    map<pii,int> k; 
    map<pii,int> mid; 
    for(int i=1;i<n;i++) 
    {
        for (int j=i+1;j<=n;j++) 
        {   
            if(i == j) continue;
            int dx = p[i].x - p[j].x;
            int dy = p[i].y - p[j].y;
            int d = __gcd(abs(dx),abs(dy));
            if(d != 0)
            {
                dx /= d;
                dy /= d;
            }
            // 统一方向变量形式，dx永远是正数
            if(dx < 0 or (dx == 0 and dy < 0))
            {
                dx = -dx;
                dy = -dy;
            }
            k[{dx,dy}]++;
            int mx = p[i].x + p[j].x;
            int my = p[i].y + p[j].y;
            mid[{mx,my}]++; // 记录中点
        }
    }
    int ans = 0;
    for (auto x : k) 
    {
        int cnt = x.second;
        ans += cnt * (cnt - 1) / 2;
    }
    int minus = 0;
    // 有多少平行四边形要减去一次，因为答案中多算了一次
    for (auto x : mid) 
    {
        int cnt = x.second;
        minus += cnt * (cnt - 1) / 2;
    }
    cout << ans - minus << endl;
}