docker_space/frontier_cs_4/local_test12.cpp

#include <bits/stdc++.h>
using namespace std;

struct TestCase {
    int n; long long k;
    vector<vector<long long>> A; long long answer;
};
mt19937_64 rng_gen(42);

TestCase gen_matrix(int n, long long k, function<long long(int,int)> valfn) {
    TestCase tc; tc.n = n; tc.k = k;
    tc.A.assign(n+1, vector<long long>(n+1, 0));
    vector<long long> all;
    for (int i = 1; i <= n; i++) for (int j = 1; j <= n; j++) { tc.A[i][j] = valfn(i, j); all.push_back(tc.A[i][j]); }
    sort(all.begin(), all.end()); tc.answer = all[k-1]; return tc;
}
TestCase gen_multiplicative(int n, long long k) { return gen_matrix(n, k, [](int i, int j) -> long long { return (long long)i * j; }); }
TestCase gen_shifted(int n, long long k) { return gen_matrix(n, k, [n](int i, int j) -> long long { return (long long)(i + n) * (j + n); }); }
TestCase gen_additive(int n, long long k) { return gen_matrix(n, k, [](int i, int j) -> long long { return i + j; }); }
TestCase gen_random_sorted(int n, long long k) {
    TestCase tc; tc.n = n; tc.k = k;
    tc.A.assign(n+1, vector<long long>(n+1, 0));
    for (int i = 1; i <= n; i++) for (int j = 1; j <= n; j++) tc.A[i][j] = (long long)i * 1000000 + (long long)j * 1000 + (rng_gen() % 500);
    for (int i = 1; i <= n; i++) for (int j = 2; j <= n; j++) tc.A[i][j] = max(tc.A[i][j], tc.A[i][j-1]);
    for (int j = 1; j <= n; j++) for (int i = 2; i <= n; i++) tc.A[i][j] = max(tc.A[i][j], tc.A[i-1][j]);
    vector<long long> all;
    for (int i = 1; i <= n; i++) for (int j = 1; j <= n; j++) all.push_back(tc.A[i][j]);
    sort(all.begin(), all.end()); tc.answer = all[k-1]; return tc;
}

struct Solver {
    const TestCase& tc;
    int query_count;
    vector<long long> memo;
    int n;

    Solver(const TestCase& t) : tc(t), query_count(0), n(t.n) { memo.assign(2002 * 2002, -1); }

    long long do_query(int r, int c) {
        int key = r * 2001 + c;
        if (memo[key] != -1) return memo[key];
        query_count++;
        memo[key] = tc.A[r][c];
        return memo[key];
    }

    long long solve() {
        long long k = tc.k;
        long long N2 = (long long)n * n;
        if (n == 1) return do_query(1, 1);

        long long heap_k = min(k, N2 - k + 1);
        if (heap_k + n <= 24000) {
            if (k <= N2 - k + 1) {
                priority_queue<tuple<long long, int, int>, vector<tuple<long long, int, int>>, greater<>> pq;
                vector<vector<bool>> vis(n + 1, vector<bool>(n + 1, false));
                pq.emplace(do_query(1, 1), 1, 1); vis[1][1] = true;
                long long result = -1;
                for (long long i = 0; i < k; i++) {
                    auto [v, r, c] = pq.top(); pq.pop(); result = v;
                    if (r+1<=n && !vis[r+1][c]) { vis[r+1][c]=true; pq.emplace(do_query(r+1,c),r+1,c); }
                    if (c+1<=n && !vis[r][c+1]) { vis[r][c+1]=true; pq.emplace(do_query(r,c+1),r,c+1); }
                }
                return result;
            } else {
                long long kk = N2 - k + 1;
                priority_queue<tuple<long long, int, int>> pq;
                vector<vector<bool>> vis(n + 1, vector<bool>(n + 1, false));
                pq.emplace(do_query(n,n),n,n); vis[n][n]=true;
                long long result = -1;
                for (long long i = 0; i < kk; i++) {
                    auto [v, r, c] = pq.top(); pq.pop(); result = v;
                    if (r-1>=1 && !vis[r-1][c]) { vis[r-1][c]=true; pq.emplace(do_query(r-1,c),r-1,c); }
                    if (c-1>=1 && !vis[r][c-1]) { vis[r][c-1]=true; pq.emplace(do_query(r,c-1),r,c-1); }
                }
                return result;
            }
        }

        // Strategy: query a strategic row entirely, use its sorted values as pivot candidates,
        // then binary search within these values using staircase walks.

        // Pick the row at position ceil(k/n) - this row's values span the likely answer range.
        // Actually, for the first pass, query the row at index ceil(k/n).
        // Its values range from a[r][1] to a[r][n], and the k-th element should be
        // somewhere in or near this range.

        vector<int> jLo(n + 1, 0), jHi(n + 1, n);
        long long cLo = 0, cHi = N2;

        // Query a strategic row
        int pivot_row = max(1, min(n, (int)((k + n - 1) / n)));
        // Query the entire row
        vector<long long> row_vals(n + 1);
        for (int j = 1; j <= n; j++) {
            row_vals[j] = do_query(pivot_row, j);
        }
        // Row values are sorted (by matrix property)

        // Binary search within this row's values for the right pivot
        int lo_idx = 1, hi_idx = n;
        while (lo_idx <= hi_idx && cHi - cLo > 0) {
            int mid_idx = (lo_idx + hi_idx) / 2;
            long long pivot = row_vals[mid_idx];

            // Staircase walk to count <= pivot
            vector<int> cutoff(n + 1, 0);
            long long cnt = 0;
            int j = jHi[1];
            for (int i = 1; i <= n; i++) {
                int lo_j = jLo[i];
                int hi_j = jHi[i];
                if (hi_j <= lo_j) { cutoff[i] = lo_j; cnt += lo_j; continue; }
                if (j > hi_j) j = hi_j;
                while (j > lo_j && do_query(i, j) > pivot) j--;
                if (j > lo_j) { cutoff[i] = j; cnt += j; }
                else { cutoff[i] = lo_j; cnt += lo_j; }
            }

            if (cnt >= k) {
                jHi = cutoff;
                cHi = cnt;
                hi_idx = mid_idx - 1;
            } else {
                jLo = cutoff;
                cLo = cnt;
                lo_idx = mid_idx + 1;
            }

            // Check if we can enumerate
            long long W = cHi - cLo;
            long long budget = 49500 - query_count;
            if (W <= budget) break;
        }

        // If still too large, query another row and refine further
        // Find a row with the most remaining candidates
        while (true) {
            long long W = cHi - cLo;
            long long budget = 49500 - query_count;
            if (W <= budget) break;
            if (budget < 2 * n + 100) break; // can't afford more walks

            // Pick a new pivot row: the row with maximum width in current band
            int best_row = -1, best_width = 0;
            for (int i = 1; i <= n; i++) {
                int w = jHi[i] - jLo[i];
                if (w > best_width) { best_width = w; best_row = i; }
            }
            if (best_row == -1 || best_width == 0) break;

            // Query the entire active segment of this row
            for (int j = jLo[best_row] + 1; j <= jHi[best_row]; j++)
                do_query(best_row, j);

            // Binary search within this row's active segment
            int lo_j = jLo[best_row] + 1, hi_j = jHi[best_row];
            // Find the value at the right quantile
            long long need_rank = k - cLo; // rank within current band
            double frac = (double)need_rank / W;
            int target_col = lo_j + (int)(frac * (hi_j - lo_j));
            target_col = max(lo_j, min(hi_j, target_col));
            long long pivot = do_query(best_row, target_col);

            // Staircase walk
            vector<int> cutoff(n + 1, 0);
            long long cnt = 0;
            int j = jHi[1];
            for (int i = 1; i <= n; i++) {
                int lo_jj = jLo[i], hi_jj = jHi[i];
                if (hi_jj <= lo_jj) { cutoff[i] = lo_jj; cnt += lo_jj; continue; }
                if (j > hi_jj) j = hi_jj;
                while (j > lo_jj && do_query(i, j) > pivot) j--;
                if (j > lo_jj) { cutoff[i] = j; cnt += j; }
                else { cutoff[i] = lo_jj; cnt += lo_jj; }
            }

            if (cnt >= k) { jHi = cutoff; cHi = cnt; }
            else { jLo = cutoff; cLo = cnt; }
        }

        // Enumerate remaining
        long long W = cHi - cLo;
        long long rank = k - cLo;
        if (W <= 0) return do_query(1, 1); // shouldn't happen

        vector<long long> cand;
        cand.reserve((size_t)W);
        for (int i = 1; i <= n; i++) {
            for (int j = jLo[i] + 1; j <= jHi[i]; j++)
                cand.push_back(do_query(i, j));
        }

        if (rank <= 0 || cand.empty()) return do_query(1, 1);
        if (rank > (long long)cand.size()) return do_query(n, n);
        nth_element(cand.begin(), cand.begin() + (rank - 1), cand.end());
        return cand[rank - 1];
    }
};

int main() {
    struct TestDef { string name; function<TestCase()> gen; };
    vector<TestDef> tests;
    tests.push_back({"additive n=100 k=5000", []{ return gen_additive(100, 5000); }});
    tests.push_back({"mult n=100 k=5000", []{ return gen_multiplicative(100, 5000); }});
    tests.push_back({"additive n=500 k=125000", []{ return gen_additive(500, 125000); }});
    tests.push_back({"mult n=500 k=125000", []{ return gen_multiplicative(500, 125000); }});
    tests.push_back({"random n=500 k=125000", []{ return gen_random_sorted(500, 125000); }});
    tests.push_back({"additive n=2000 k=2000000", []{ return gen_additive(2000, 2000000); }});
    tests.push_back({"mult n=2000 k=2000000", []{ return gen_multiplicative(2000, 2000000); }});
    tests.push_back({"shifted n=2000 k=2000000", []{ return gen_shifted(2000, 2000000); }});
    tests.push_back({"random n=2000 k=2000000", []{ return gen_random_sorted(2000, 2000000); }});
    tests.push_back({"mult n=2000 k=100000", []{ return gen_multiplicative(2000, 100000); }});
    tests.push_back({"mult n=2000 k=3900000", []{ return gen_multiplicative(2000, 3900000); }});

    for (auto& t : tests) {
        auto tc = t.gen();
        Solver s(tc);
        long long result = s.solve();
        bool correct = (result == tc.answer);
        int used = s.query_count;
        double score = !correct ? 0.0 : (used <= tc.n ? 1.0 : (used >= 50000 ? 0.0 : (50000.0 - used) / (50000.0 - tc.n)));
        printf("%-45s q=%6d %s score=%.4f\n", t.name.c_str(), used, correct ? "OK" : "WRONG", score);
    }
}