CF 207C1 - Game with Two Trees

CF 207C1 - Game with Two Trees

Rating: 2100
Tags: -
Solve time: 3m 32s
Verified: no

Solution

Problem Understanding

We are asked to maintain two dynamically growing rooted trees with labeled edges, and after each operation, compute the number of "good combinations." A good combination consists of a vertex from the first tree and a pair of vertices in the second tree such that the string obtained by walking backward from the first tree vertex to the root matches the string obtained by walking forward along a path in the second tree. Operations progressively add a new child to a specified vertex with a specified character, and we must update the count of all good combinations after each operation.

The input consists of up to $10^5$ operations. A naive approach that explicitly compares backward paths from every node in the first tree to all forward paths in the second tree would take roughly $O(n^3)$ time in the worst case, which is impossible. Thus, we need a solution that updates counts incrementally in near-linear time.

Edge cases arise when a newly added node immediately forms multiple new matches because it completes multiple paths in the second tree simultaneously. For example, adding a child with the same character to multiple nodes can match different backward paths in the first tree. Careless implementations might miss these multiple matches or double count them incorrectly.

Approaches

The brute-force solution builds both trees explicitly. After each operation, it enumerates every vertex in tree one and every possible path in tree two to compare strings. This approach is correct but clearly too slow: for $n = 10^5$ nodes, the number of paths in a tree is quadratic in the number of nodes, giving roughly $O(n^3)$ operations for each update.

The key insight is that backward paths in the first tree and forward paths in the second tree can be represented as strings, but comparing strings directly is slow. Instead, we can assign unique hash values to each path using a rolling hash (or trie). This transforms the string comparison into an integer comparison. Each vertex's backward path hash in tree one can be stored and updated incrementally. For tree two, we maintain a trie of forward paths, storing counts of how many vertices correspond to each prefix. When a new vertex is added, we update the trie and increment the global count of good combinations by the number of matching backward paths from tree one.

This reduces complexity because each operation processes only the new vertex and its immediate connections. The hash/trie representation ensures that string matching and counting can be done in $O(\log n)$ or $O(1)$ amortized per update depending on the implementation.

Algorithm Walkthrough

1. Maintain for tree one an array backward_hash[i] representing the hash of the backward path from vertex i to the root. For a new child v_new with parent v and character c, compute backward_hash[v_new] = backward_hash[v] * P + ord(c), where P is a suitable base for rolling hash.
2. Maintain for tree two a trie structure forward_trie where each node contains a count of vertices that correspond to the string path from the root to that trie node. Each edge is labeled by a character. Initially, the root node has count 1 for the empty string.
3. For each operation on tree two, traverse the trie along the parent path and extend it by the new character. Update counts at the new trie node.
4. Maintain a global map hash_count that counts how many vertices in tree one have each backward hash. When a vertex is added in tree one, increment the count for its backward_hash.
5. To count new good combinations after an operation, if the operation was on tree one, check the backward hash of the new vertex and add forward_trie_count[hash] to the total count. If the operation was on tree two, traverse backward from the new vertex to root (or forward along trie) and for each prefix hash in the trie, add hash_count[prefix_hash] to the total.
6. Output the total good combinations after each operation.

This ensures incremental updates in $O(1)$ amortized time per operation because each vertex is processed once and only contributes via its hash to the total count.

Why it works: The algorithm maintains the invariant that hash_count[h] contains the number of backward paths in tree one matching hash h and forward_trie contains the number of forward paths in tree two for each string prefix. The number of good combinations after each operation is exactly the sum over all matching hashes, guaranteeing correctness.

Python Solution

import sys
input = sys.stdin.readline

class TrieNode:
    def __init__(self):
        self.children = {}
        self.count = 0

def solve():
    n = int(input())
    P = 31
    MOD = 10**9 + 7
    
    # Tree 1
    tree1 = [0]  # parent indices
    hash1 = [0]  # backward hash to root
    hash_count = {0: 1}  # empty root
    
    # Tree 2
    tree2 = [0]
    trie_root = TrieNode()
    trie_nodes = [trie_root]
    
    total_good = 0
    
    for _ in range(n):
        t, v, c = input().split()
        t = int(t)
        v = int(v) - 1
        c_val = ord(c) - ord('a') + 1
        
        if t == 1:
            parent_hash = hash1[v]
            new_hash = (parent_hash * P + c_val) % MOD
            hash1.append(new_hash)
            tree1.append(v)
            hash_count[new_hash] = hash_count.get(new_hash, 0)
            
            # Count new good combinations
            total_good += trie_nodes[v].count if v < len(trie_nodes) else 0
        else:
            parent_node = trie_nodes[v]
            if c not in parent_node.children:
                parent_node.children[c] = TrieNode()
            new_node = parent_node.children[c]
            new_node.count = 1
            trie_nodes.append(new_node)
            tree2.append(v)
            
            # Count new good combinations