Editorial

Let’s add the names of the cities from each test case to a set of strings. The number of visited cities will be equal to the size of this set.

Algorithm realization

Declare a string $w or d$ and a set of strings $s$ .

string word;
set<string> s;

Read the number of test cases $t es t s$ and process each of the tests sequentially.

cin >> tests;
while(tests--)
{

Read the number of cities $n$ . Clear the set $s$ .

  cin >> n;
  s.clear();

Read the city names and add them to the set $s$ .

  while(n--)
  {
    cin >> word;
    s.insert(word);
  }

Print the number of distinct visited cities.

  cout << s.size() << endl;
}

Java realization

import java.util.*;
 
public class Main
{
  public static void main(String[] args)
  {
    Scanner con = new Scanner(System.in);
    TreeSet<String> s = new TreeSet<String>();
 
    int tests = con.nextInt();
    for(int i = 0; i < tests; i++)
    {
      int n = con.nextInt();
      s.clear();
      for(int j = 0; j < n; j++)
      {
        String word = con.next();
        s.add(word);
      }
      System.out.println(s.size());      
    }
    con.close();
  }
}

Python realization

Read the number of test cases $t es t s$ and process each of the tests sequentially.

tests = int(input())
for _ in range(tests):

Read the number of cities $n$ . Clear the set $s$ .

  n = int(input())
  s = set({})

Read the city names and add them to the set $s$ .

  for _ in range(n):
    s.add(input())

Print the number of distinct visited cities.

  print(len(s))