June 17, 2020
解説のように、でも解けるが、座標圧縮 + 二次元累積和で事前に長方形に含まれる点の個数を計算しておくこともできる。
int main() { i64 N, K; cin >> N >> K; using P = pair<i64, i64>; vector<P> point(N); map<i64, i64> mpX, mpY; REP(i, N) { cin >> point[i].first >> point[i].second; mpX[point[i].first] = 0; mpY[point[i].second] = 0; } vector<i64> xvals; vector<i64> yvals; for (auto &p : mpX) { p.second = xvals.size(); xvals.pb(p.first); } for (auto &p : mpY) { p.second = yvals.size(); yvals.pb(p.first); } i64 H = yvals.size() * 2; i64 W = xvals.size() * 2; vector<vector<i64>> G(H, vector<i64>(W, 0)); REP(i, N) { i64 y = mpY[point[i].second] * 2; i64 x = mpX[point[i].first] * 2; G[y][x] = 1; } vector<vector<i64>> C(H + 1, vector<i64>(W + 1, 0)); REP(y, H) { REP(x, W) { C[y + 1][x + 1] = C[y][x + 1] + C[y + 1][x] - C[y][x] + G[y][x]; } } i64 mins = numeric_limits<i64>::max(); REP(i, N) REP(j, N) REP(k, N) REP(l, N) { i64 minx, miny, maxx, maxy; minx = maxx = point[i].first; miny = maxy = point[i].second; auto update = [&](i64 idx) { chmin(minx, point[idx].first); chmax(maxx, point[idx].first); chmin(miny, point[idx].second); chmax(maxy, point[idx].second); }; update(j), update(k), update(l); i64 clx, cly, crx, cry; clx = mpX[minx] * 2; cly = mpY[miny] * 2; crx = mpX[maxx] * 2 + 1; cry = mpY[maxy] * 2 + 1; i64 cnt = C[cry][crx] - C[cry][clx] - C[cly][crx] + C[cly][clx]; if (cnt >= K) { i64 dx = abs(maxx - minx); i64 dy = abs(maxy - miny); if (dx * dy > 0) chmin(mins, dx * dy); } } cout << mins << endl; return 0; }