## Intuition The problem involves checking if a given permutation of nodes in a perfect binary tree can represent a valid DFS (Depth-First Search) traversal order. A perfect binary tree is a complete tree where all levels are fully filled. The DFS traversal of such a tree is not unique, but it must respect the parent-child relationships. The key insight is that for a permutation to be a valid DFS order, whenever a node appears in the permutation, all its ancestors must have appeared before it. This is because DFS visits nodes along a path from the root to a leaf before backtracking. ## Approach 1. **Tree Representation**: Given the nature of a perfect binary tree, we can derive the parent-child relationships directly. Each node `i` has a parent at `i/2`. 2. **DFS Validity Check**: To check if a permutation is a valid DFS order, we maintain a set of visited nodes. As we iterate through the permutation, we ensure that each node's parent has been visited before the node itself. 3. **Handling Queries**: For each query, we swap two elements in the permutation and check if the new permutation is a valid DFS order using the above method. 4. **Efficiency**: The check for DFS validity is linear with respect to the number of nodes, making it feasible to handle the constraints. ## Complexity - **Time**: The DFS validity check is `O(n)` per query, and with `q` queries, the complexity per test case is `O(n * q)`. - **Space**: `O(n)` for storing the visited nodes. ## Solution (C++) ```cpp #include using namespace std; bool isValidDFS(const vector& p, int n) { vector visited(n + 1, false); for (int i = 0; i < n; ++i) { int node = p[i]; int parent = node / 2; if (parent > 0 && !visited[parent]) { return false; // Parent must be visited before the node } visited[node] = true; } return true; } int main() { ios::sync_with_stdio(false); cin.tie(nullptr); int t; cin >> t; while (t--) { int n, q; cin >> n >> q; vector parent(n + 1); for (int i = 2; i <= n; ++i) { cin >> parent[i]; // Read parent, but not needed due to perfect binary tree property } vector p(n); for (int i = 0; i < n; ++i) { cin >> p[i]; } while (q--) { int x, y; cin >> x >> y; swap(p[x - 1], p[y - 1]); if (isValidDFS(p, n)) { cout << "YES\n"; } else { cout << "NO\n"; } } } return 0; } ``` This solution reads the input, processes each query by swapping the specified elements in the permutation, and checks if the resulting permutation is a valid DFS order. The use of a `visited` array helps ensure that all ancestors of a node appear before the node itself in the permutation.

DFS Checker (Easy Version)

Problem Details