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| using namespace std; | |
| int n; | |
| long long k; | |
| vector<long long> memo; | |
| int query_count = 0; | |
| long long do_query(int r, int c) { | |
| int key = r * 2001 + c; | |
| if (memo[key] != -1) return memo[key]; | |
| cout << "QUERY " << r << " " << c << "\n"; | |
| cout.flush(); | |
| long long v; | |
| cin >> v; | |
| memo[key] = v; | |
| query_count++; | |
| return v; | |
| } | |
| void done(long long ans) { | |
| cout << "DONE " << ans << "\n"; | |
| cout.flush(); | |
| exit(0); | |
| } | |
| void solve() { | |
| long long total = (long long)n * n; | |
| if (n == 1) { done(do_query(1, 1)); return; } | |
| // Heap for extreme k | |
| long long heap_k = min(k, total - k + 1); | |
| if (heap_k + n <= 24000) { | |
| if (k <= total - 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); } | |
| } | |
| done(result); | |
| } else { | |
| long long kk = total - 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); } | |
| } | |
| done(result); | |
| } | |
| } | |
| // Single-walk quickselect: only do <= walk, no < walk | |
| // This saves one walk per iteration at the cost of not detecting exact pivot match | |
| vector<int> L(n + 1, 1), R(n + 1, n); | |
| long long k_rem = k; | |
| 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) done(do_query(i, L[i])); | |
| break; | |
| } | |
| // Heap fallback | |
| 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); | |
| } | |
| done(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); | |
| } | |
| done(get<0>(pq.top())); | |
| } | |
| // Pivot selection | |
| vector<long long> pvals; | |
| double target_frac = (double)(k_rem - 0.5) / total_cand; | |
| 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]; | |
| // Single staircase walk: bottom-to-top, left-to-right (reversed) | |
| 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; | |
| } | |
| if (cle >= k_rem) { | |
| // Answer is <= pivot. Narrow right. | |
| for (int i : active) R[i] = min(R[i], p_le[i]); | |
| } else { | |
| // Answer is > pivot. Narrow left. | |
| k_rem -= cle; | |
| for (int i : active) L[i] = max(L[i], p_le[i] + 1); | |
| } | |
| } | |
| } | |
| int main() { | |
| ios_base::sync_with_stdio(false); | |
| cin.tie(nullptr); | |
| cin >> n >> k; | |
| memo.assign(2002 * 2002, -1); | |
| solve(); | |
| return 0; | |
| } | |