简单题。
首先肯定是要枚举梯形其中一条底边的两个端点的。那么另一条底边除了斜率与这条边相等，两个端点的距离要分别与这条底边两个端点的距离相等。
发现这个十分不好做，考虑一个梯形是等腰梯形的一个充要条件。不难想到，连接两底中点，这条线段垂直于两条底边。于是除了要满足斜率相等外，还要满足两个点的中点连线的斜率为底边斜率的负倒数。要限定 \(\frac{y_1 - y_2}{x_1 - x_2} = k\)，移项得到 \(y_1 - k x_1 = y_2 - k x_2\)。于是只需限定这一项相等即可。实现时可以用 map。
注意一些细节，为了避免精度问题可以自己手写分数类，而且要特殊处理底边为水平线或竖直线的情况，两底还不能共线（可以把重合的直线放一起处理）。
时间复杂度 \(O(n^2 \log n)\)，常数有点大。
code
// Problem: G - Isosceles Trapezium
// Contest: AtCoder - AtCoder Beginner Contest 220
// URL: https://atcoder.jp/contests/abc220/tasks/abc220_g
// Memory Limit: 1024 MB
// Time Limit: 3000 ms
//
// Powered by CP Editor (https://cpeditor.org)
#include <bits/stdc++.h>
#define pb emplace_back
#define fst first
#define scd second
#define mems(a, x) memset((a), (x), sizeof(a))
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef double db;
typedef long double ldb;
typedef pair<ll, ll> pii;
const int maxn = 1010;
struct frac {
ll x, y;
frac(ll a = 0, ll b = 1) {
if (b < 0) {
a = -a;
b = -b;
}
x = a;
y = b;
}
};
inline bool operator < (const frac &a, const frac &b) {
return a.x * b.y < a.y * b.x;
}
inline bool operator == (const frac &a, const frac &b) {
return a.x * b.y == a.y * b.x;
}
inline frac operator + (const frac &a, const frac &b) {
return frac(a.x * b.y + a.y * b.x, a.y * b.y);
}
inline frac operator - (const frac &a, const frac &b) {
return frac(a.x * b.y - a.y * b.x, a.y * b.y);
}
inline frac operator * (const frac &a, const frac &b) {
return frac(a.x * b.x, a.y * b.y);
}
ll n;
struct node {
ll x, y, z;
} a[maxn];
inline frac getb(int i, int j) {
frac k(a[i].y - a[j].y, a[i].x - a[j].x);
return frac(a[i].y) - k * frac(a[i].x);
}
void solve() {
scanf("%lld", &n);
for (int i = 1; i <= n; ++i) {
scanf("%lld%lld%lld", &a[i].x, &a[i].y, &a[i].z);
}
ll ans = -1;
sort(a + 1, a + n + 1, [&](node a, node b) {
return a.x < b.x;
});
map<frac, ll> M;
for (int i = 1, j = 1; i <= n; i = (++j)) {
while (j < n && a[j + 1].x == a[i].x) {
++j;
}
for (int k = i; k <= j; ++k) {
for (int l = k + 1; l <= j; ++l) {
frac mid(a[k].y + a[l].y, 2);
if (M.find(mid) != M.end()) {
ans = max(ans, M[mid] + a[k].z + a[l].z);
}
}
}
for (int k = i; k <= j; ++k) {
for (int l = k + 1; l <= j; ++l) {
frac mid(a[k].y + a[l].y, 2);
M[mid] = max(M[mid], a[k].z + a[l].z);
}
}
}
M.clear();
sort(a + 1, a + n + 1, [&](node a, node b) {
return a.y < b.y;
});
for (int i = 1, j = 1; i <= n; i = (++j)) {
while (j < n && a[j + 1].y == a[i].y) {
++j;
}
for (int k = i; k <= j; ++k) {
for (int l = k + 1; l <= j; ++l) {
frac mid(a[k].x + a[l].x, 2);
if (M.find(mid) != M.end()) {
ans = max(ans, M[mid] + a[k].z + a[l].z);
}
}
}
for (int k = i; k <= j; ++k) {
for (int l = k + 1; l <= j; ++l) {
frac mid(a[k].x + a[l].x, 2);
M[mid] = max(M[mid], a[k].z + a[l].z);
}
}
}
map< frac, vector<pii> > mp;
for (int i = 1; i <= n; ++i) {
for (int j = i + 1; j <= n; ++j) {
if (a[i].x == a[j].x || a[i].y == a[j].y) {
continue;
}
frac k1(a[j].y - a[i].y, a[j].x - a[i].x);
mp[k1].pb(i, j);
}
}
for (auto pp : mp) {
frac k1 = pp.fst;
map<frac, ll> M;
vector<pii> vc = pp.scd;
sort(vc.begin(), vc.end(), [&](pii a, pii b) {
return getb(a.fst, a.scd) < getb(b.fst, b.scd);
});
int len = (int)vc.size();
for (int i = 0, j = 0; i < len; i = (++j)) {
while (j + 1 < len && getb(vc[j + 1].fst, vc[j + 1].scd) == getb(vc[i].fst, vc[i].scd)) {
++j;
}
for (int k = i; k <= j; ++k) {
frac mx(a[vc[k].fst].x + a[vc[k].scd].x, 2), my(a[vc[k].fst].y + a[vc[k].scd].y, 2);
frac k2(-k1.y, k1.x);
frac fr = my - mx * k2;
if (M.find(fr) != M.end()) {
ans = max(ans, M[fr] + a[vc[k].fst].z + a[vc[k].scd].z);
}
}
for (int k = i; k <= j; ++k) {
frac mx(a[vc[k].fst].x + a[vc[k].scd].x, 2), my(a[vc[k].fst].y + a[vc[k].scd].y, 2);
frac k2(-k1.y, k1.x);
frac fr = my - mx * k2;
M[fr] = max(M[fr], a[vc[k].fst].z + a[vc[k].scd].z);
}
}
}
printf("%lld\n", ans);
}
int main() {
int T = 1;
// scanf("%d", &T);
while (T--) {
solve();
}
return 0;
}