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AtCoder ABC 074 D

D - Restoring Road Network

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;
}