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| 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]; | |
| } | |
| // Staircase walk with L[i]/R[i] bounds, returns cle and p_le | |
| pair<long long, vector<int>> walk_le(long long pivot, const vector<int>& active, const vector<int>& L, const vector<int>& R) { | |
| int na = active.size(); | |
| vector<int> p_le(n + 1, 0); | |
| int j = 0; | |
| for (int idx = na - 1; idx >= 0; idx--) { | |
| int i = active[idx]; | |
| j = max(j, L[i]); | |
| while (j <= R[i] && do_query(i, j) <= pivot) j++; | |
| p_le[i] = j - 1; | |
| } | |
| long long cle = 0; | |
| for (int i : active) { | |
| int rl = min(p_le[i], R[i]); | |
| if (rl >= L[i]) cle += rl - L[i] + 1; | |
| } | |
| return {cle, p_le}; | |
| } | |
| 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; | |
| } | |
| } | |
| vector<int> L(n + 1, 1), R(n + 1, n); | |
| long long k_rem = k; | |
| // Keep track of value bounds for binary search refinement | |
| long long val_lo = -1, val_hi = -1; // will be set after first walk | |
| long long cnt_lo = 0, cnt_hi = -1; | |
| for (int iter = 0; iter < 100; iter++) { | |
| vector<int> active; | |
| long long total_cand = 0; | |
| for (int i = 1; i <= n; i++) { | |
| if (L[i] <= R[i]) { active.push_back(i); total_cand += R[i] - L[i] + 1; } | |
| } | |
| int na = active.size(); | |
| if (total_cand == 0) break; | |
| if (total_cand == 1) { for (int i : active) return do_query(i, L[i]); break; } | |
| long long budget = 49500 - query_count; | |
| if (k_rem + na <= budget) { | |
| priority_queue<tuple<long long, int, int>, vector<tuple<long long, int, int>>, greater<>> pq; | |
| for (int i : active) pq.emplace(do_query(i, L[i]), i, L[i]); | |
| for (long long t = 1; t < k_rem; t++) { | |
| auto [v, r, c] = pq.top(); pq.pop(); | |
| if (c + 1 <= R[r]) pq.emplace(do_query(r, c + 1), r, c + 1); | |
| } | |
| return get<0>(pq.top()); | |
| } | |
| long long rev_k = total_cand - k_rem + 1; | |
| if (rev_k + na <= budget) { | |
| priority_queue<tuple<long long, int, int>> pq; | |
| for (int i : active) pq.emplace(do_query(i, R[i]), i, R[i]); | |
| for (long long t = 1; t < rev_k; t++) { | |
| auto [v, r, c] = pq.top(); pq.pop(); | |
| if (c - 1 >= L[r]) pq.emplace(do_query(r, c - 1), r, c - 1); | |
| } | |
| return get<0>(pq.top()); | |
| } | |
| // Pivot selection (original method) | |
| double target_frac = (double)(k_rem - 0.5) / total_cand; | |
| vector<long long> pvals; | |
| int sample_n = max(1, min(na, (int)ceil(sqrt((double)na) * 4))); | |
| int step = max(1, na / sample_n); | |
| for (int idx = 0; idx < na; idx += step) { | |
| int i = active[idx]; | |
| int width = R[i] - L[i] + 1; | |
| int col = L[i] + (int)(target_frac * width); | |
| col = max(L[i], min(R[i], col)); | |
| pvals.push_back(do_query(i, col)); | |
| } | |
| sort(pvals.begin(), pvals.end()); | |
| long long pivot = pvals[pvals.size() / 2]; | |
| // Staircase walk | |
| auto [cle, p_le] = walk_le(pivot, active, L, R); | |
| // Track value bounds | |
| if (cle >= k_rem) { | |
| val_hi = pivot; | |
| cnt_hi = cle; | |
| } else { | |
| val_lo = pivot; | |
| cnt_lo = cle; // note: this is cle within current L/R bounds | |
| } | |
| // Apply the partition as usual | |
| if (cle >= k_rem) { | |
| for (int i : active) R[i] = min(R[i], p_le[i]); | |
| } else { | |
| k_rem -= cle; | |
| for (int i : active) L[i] = max(L[i], p_le[i] + 1); | |
| } | |
| // After partition, check if we can do a VALUE-BASED binary search refinement | |
| // If we have both val_lo and val_hi, and they're close enough, binary search between them | |
| // Each binary search step is another staircase walk (O(n) queries) | |
| // This is useful when the quickselect gets stuck with bad splits | |
| if (val_lo >= 0 && val_hi >= 0 && val_lo < val_hi) { | |
| // Re-check remaining candidates | |
| long long new_total = 0; | |
| for (int i = 1; i <= n; i++) { | |
| if (L[i] <= R[i]) new_total += R[i] - L[i] + 1; | |
| } | |
| // If we haven't narrowed enough and budget allows, do one more walk with interpolated pivot | |
| if (new_total > budget - 2000 && budget > 4 * n) { | |
| long long mid_val = val_lo + (val_hi - val_lo) / 2; | |
| if (mid_val > val_lo && mid_val < val_hi) { | |
| vector<int> active2; | |
| long long tc2 = 0; | |
| for (int i = 1; i <= n; i++) { | |
| if (L[i] <= R[i]) { active2.push_back(i); tc2 += R[i] - L[i] + 1; } | |
| } | |
| if (!active2.empty()) { | |
| auto [cle2, p_le2] = walk_le(mid_val, active2, L, R); | |
| if (cle2 >= k_rem) { | |
| for (int i : active2) R[i] = min(R[i], p_le2[i]); | |
| val_hi = mid_val; | |
| } else { | |
| k_rem -= cle2; | |
| for (int i : active2) L[i] = max(L[i], p_le2[i] + 1); | |
| val_lo = mid_val; | |
| } | |
| } | |
| } | |
| } | |
| } | |
| } | |
| return -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); | |
| } | |
| } | |