2022/finals/alphabet_adventuring.md

Alphonse is assembling an *alphabet tree* and arranging some adventures along the way.

An alphabet tree is an unrooted tree with \(N\) nodes (numbered from \(1\) to \(N\)) and \(N - 1\) edges. Initially, the \(i\)th edge connects nodes \(A_i\) and \(B_i\) in both directions, and is labeled with a uppercase letter \(C_i\). Two edges incident to a common node are always labeled with different letters.

Alphonse has \(Q\) events to process, the \(i\)th of which is one of two types:

* `1` \(U_i\) \(L_i\): Add a new node to the tree by connecting it to node \(U_i\) with a new edge labeled with uppercase letter \(L_i\). Newly added nodes are numbered with integers starting from \(N + 1\) in the order they're added.
* `2` \(U_i\) \(K_i\) \(S_i\): Print the final node Alphonse will end up at if he:
  - Starts a journey at node \(U_i\).
  - Repeatedly traverses a previously untraversed edge (on this journey). If his current node has multiple untraversed edges, he picks the edge labeled with the letter that comes earliest in the string \(S_i\).
  - Ends the journey once there are no more untraversed edges at the current node, or \(K_i\) edges have been traversed on the journey.

Please help Alphonse determine where each journey will take him.


# Constraints

\(1 \le T \le 20\)
\(2 \le N \le 300{,}000\)
\(1 \le Q \le 300{,}000\)
\(1 \le A_i, B_i \le N\)
\(A_i \ne B_i\)
\(C_i, L_i, S_{i,j} \in \{\)`'A'`, \(\ldots\), `'Z'`\(\}\)
\(1 \le U_i, K_i \le 600{,}000\)

The sum of \(N\) over all test cases is at most \(1{,}100{,}000\).
The sum of \(Q\) over all test cases is at most \(1{,}100{,}000\).

For each event, it is guaranteed that:
- \(U_i\) is a valid node in the tree at the time of the event.
- \(L_i\) is different from all existing labels of edges incident to \(U_i\) at the time of the event.
- \(S_i\)'s letters are distinct, and are a superset of all edge labels in the tree at the time of the event.


# Input Format

Input begins with a single integer \(T\), the number of test cases. For each test case, there is first a line containing a single integer \(N\). Then, \(N - 1\) lines follow, the \(i\)th of which contains space-separated integers \(A_i\) and \(B_i\), followed by a space, followed by \(C_i\). Then, there is a line containing the single integer \(Q\). Then, \(Q\) lines follow, the \(i\)th of which is either `1` \(U_i\) \(L_i\) or `2` \(U_i\) \(K_i\) \(S_i\).


# Output Format

For the \(i\)th test case, print a single line containing `"Case #i: "`, followed by space-separated integers, the answers to all type-2 events.


# Sample Explanation

{{PHOTO_ID:1175696903384597|WIDTH:550}}

The first sample case's alphabet tree is depicted above, and yields the following journeys:

- Event \(1\) traverses \(1 \stackrel{\texttt{M}}{\longrightarrow} 2 \stackrel{\texttt{E}}{\longrightarrow} 6\) (ended after no more edges from node \(6\))
- Event \(2\) traverses \(3 \stackrel{\texttt{E}}{\longrightarrow} 1 \stackrel{\texttt{T}}{\longrightarrow} 4 \stackrel{\texttt{A}}{\longrightarrow} 9\) (ended after \(K_2 = 3\) steps)
- Event \(3\) traverses \(9 \stackrel{\texttt{A}}{\longrightarrow} 4 \stackrel{\texttt{T}}{\longrightarrow} 1 \stackrel{\texttt{M}}{\longrightarrow} 2\) (ended after \(K_3 = 3\) steps)
- Event \(4\) traverses \(8 \stackrel{\texttt{M}}{\longrightarrow} 3 \stackrel{\texttt{E}}{\longrightarrow} 1 \stackrel{\texttt{T}}{\longrightarrow} 4 \stackrel{\texttt{A}}{\longrightarrow} 9\) (ended after no more edges from node \(9\))
- Event \(6\) traverses \(8 \stackrel{\texttt{T}}{\longrightarrow} 10\) (ended after no more edges from node \(10\))

The second and third sample cases are depicted below.

{{PHOTO_ID:1142476319774543|WIDTH:650}}

In the second case, the journeys are: 

- Event \(1\) traverses \(1 \stackrel{\texttt{P}}{\longrightarrow} 5 \stackrel{\texttt{C}}{\longrightarrow} 2\) (ended after \(K_1 = 2\) steps)
- Event \(2\) traverses \(4 \stackrel{\texttt{P}}{\longrightarrow} 2 \stackrel{\texttt{C}}{\longrightarrow} 5 \stackrel{\texttt{P}}{\longrightarrow} 1\) (ended after \(K_2 = 3\) steps)
- Event \(4\) traverses \(4 \stackrel{\texttt{P}}{\longrightarrow} 2 \stackrel{\texttt{C}}{\longrightarrow} 5 \stackrel{\texttt{U}}{\longrightarrow} 6\) (ended after \(K_4 = 3\) steps)
- Event \(5\) traverses \(3 \stackrel{\texttt{U}}{\longrightarrow} 2 \stackrel{\texttt{P}}{\longrightarrow} 4\) (ended after no more edges from node \(4\))

In the third case, the journeys are: 

- Event \(1\) traverses \(3 \stackrel{\texttt{C}}{\longrightarrow} 2 \stackrel{\texttt{A}}{\longrightarrow} 1\) (ended after \(K_1 = 2\) steps)
- Event \(2\) traverses \(4 \stackrel{\texttt{E}}{\longrightarrow} 2 \stackrel{\texttt{A}}{\longrightarrow} 1\) (ended after \(K_2 = 2\) steps)
- Event \(3\) traverses \(1 \stackrel{\texttt{A}}{\longrightarrow} 2\) (ended after \(K_3 = 1\) steps)