In order to solve this problem, you need a knowledge of tree structures and bipartiteness, and of either BFS (Breadth-First Search) or DFS (Depth-First Search). Such concepts is so frequently seen that you can refer to the materials on the Internet.

If you heard such terms for the first time, we recommend you to search on the Internet before reading this Editorial.

Kingdom of Takahashi forms a “tree structure” in a graph-theory term.

Assuming each road has length \(1\), let us paint red those town whose shortest distance to Town \(i\) is even; otherwise, let us paint them blue.

Then, due to the bipartiteness of trees, we can see that if a person is currently at a red town, then he next goes to blue town; conversely, a person who is now at a blue town will always move to a red town. Therefore, if the colors of the vertices where Takahashi and Aoki were initially were are different, then they never meet at the same town; On the other hand, if those colors are same, they never meet at a road.
Therefore, after we precalculate the color of each town, we can answer each query in an \(O(1)\) time.

To find the colors of the towns, we can use BFS or DFS.

Similar problem：https://atcoder.jp/contests/abc126/tasks/abc126_d

#include<bits/stdc++.h>
using namespace std;

int main(){
    int N, Q; cin >> N >> Q;
    vector<vector<int>> G(N);
    for(int i = 0; i < N-1; i++){
        int A, B; cin >> A >> B;
        G[A-1].push_back(B-1);
        G[B-1].push_back(A-1);
    }
    queue<int> que;
    vector<int> dis(N,-1);
    que.push(0);
    dis[0] = 0;
    while(!que.empty()){
        int t = que.front(); que.pop();
        for(int x: G[t]) if(dis[x] == -1) {
            dis[x] = dis[t] + 1;
            que.push(x);
        }
    }
    for(int i = 0; i < Q; i++){
        int A, B;cin >> A >> B;
        if(dis[A-1]%2 == dis[B-1]%2) cout << "Town" << endl;
        if(dis[A-1]%2 != dis[B-1]%2) cout << "Road" << endl;
    }
}

import queue

N,Q=map(int,input().split())
G=[[] for i in range(N)]
for i in range(N-1):
    a,b=map(int,input().split())
    G[a-1].append(b-1)
    G[b-1].append(a-1)
que=queue.Queue()
color=[-1]*N
color[0]=0
que.put(0)
while not que.empty():
    t=que.get()
    for i in G[t]:
        if color[i] == -1:
            color[i] = 1 - color[t]
            que.put(i)
for i in range(Q):
    a,b=map(int,input().split())
    if color[a-1] == color[b-1]:
        print("Town")
    else:
        print("Road")