本来以为是上回那道数正方形的原题的。结果发现不要求边与座标轴平行，于是我就naive了。然后数据结构变身计算几何一堆东西吼吼。
矩形的对角线相等且互相平分（来自初二数学书）。把所有点对的中点找出来，把所有相等且平分的放一起用单调性，有点像旋转卡(qia3)壳(qiao4)的做法。jason_yu说暴力枚举都能过，而且他跑得比我快！
计算几何也写得比较naive。调了半天啊。
毕竟我弱嘛。
#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;
typedef long long qw;
#ifdef WIN32
#define lld "%I64d"
#else
#define lld "%lld"
#endif
#define _l (qw)
const double eps = 1e-8;
const int maxn = 1509;
#define sqr(x) ((x)*(x))
#define dist(a,b) (sqr(a.x-b.x)+sqr(a.y-b.y))
#define nextele(_x_) ((_x_+1==r)?(l):(_x_+1))
struct point {
qw x, y;
point(qw x0 = 0, qw y0 = 0) {
x = x0, y = y0;
}
inline point operator =(const point& a) {
x = a. x, y = a. y;
return *this;
}
};
struct vect {
qw x, y;
inline vect(qw x0 = 0, qw y0 = 0) {
x = x0, y = y0;
}
inline vect(point a, point b) {
x = b. x - a. x, y = b. y - a. y;
}
inline qw mulx(const vect& a) const {
return x * a. y - y * a. x;
}
inline vect operator =(const vect& a) {
x = a. x, y = a. y;
return *this;
}
inline void reverse() {
x = -x, y = -y;
}
};
struct mid_point {
qw x, y, l;
vect d;
inline mid_point() {
}
inline mid_point(point a, point b) {
x = (a. x + b. x) >> 1;
y = (a. y + b. y) >> 1;
l = dist(a, b);
d = vect(b. x - x, b. y - y);
if (d. x < 0)
d. reverse();
}
inline bool operator ==(const mid_point& a) {
return x == a. x && y == a. y && l == a. l;
}
inline mid_point operator =(const mid_point& a) {
x = a. x, y = a. y, l = a. l, d = a. d;
return *this;
}
};
inline bool cmpPoint(const point& a, const point& b) {
return a. x < b. x || (a. x == b. x && a. y < b. y);
}
inline bool cmpMPoint(const mid_point& a, const mid_point& b) {
return a. x < b. x || (a. x == b. x && a. y < b. y) || \
(a. x == b. x && a. y == b. y && a. l < b. l) || \
(a. x == b. x && a. y == b. y && a. l == b. l && a. d. mulx(b. d) > 0);
}
int n, t;
qw ans;
point a[maxn];
mid_point b[maxn * maxn];
int main() {
#ifndef ONLINE_JUDGE
freopen("in.txt", "r", stdin);
#endif
ans = 0;
t = 0;
scanf("%d", &n);
for (int i = 0; i < n; ++ i) {
scanf(lld lld, &a[i]. x, &a[i]. y);
a[i]. x <<= 1;
a[i]. y <<= 1;
for (int j = 0; j < i; ++ j)
b[t ++] = mid_point(a[i], a[j]);
}
sort(b, b + t, cmpMPoint);
for (int l = 0, r; l < t; l = r) {
for (r = l + 1; r < t && b[r] == b[l]; ++ r);
if (r - l < 2)
continue;
for (int i = l, j = l; i < r; ++ i) {
if (j == i)
j = nextele(j);
while (b[i]. d. mulx(b[j]. d) < b[i]. d. mulx(b[nextele(j)]. d))
j = nextele(j);
ans = max(ans, abs(b[i]. d. mulx(b[j]. d)));
}
}
printf(lld "\n", ans >> 1);
}