Overview

This problem asks us to find the total number of pairs of players who are entered in the same event but belong to different teams.

Analysis

A naive approach would check all pairs of players (\(O(N^2)\) pairs), but since \(N\) can be up to \(2 \times 10^5\), this method would not meet the time limit.

The key observation is that matches can only occur within the same event, and players on the same team cannot compete against each other. Therefore, we need to group players by event and efficiently calculate the number of pairs of players from different teams within each event.

Algorithm

1. Group players by event number
2. For each event:
   • The total number of pairs from all \(T\) players in the event is \(\frac{T(T-1)}{2}\)
   • Subtract the number of pairs within the same team: \(\sum_{\text{each team}} \frac{c_i(c_i-1)}{2}\)
   • This gives the number of pairs between different teams within the event
3. Sum the above values across all events

This approach uses a dictionary of dictionaries (defaultdict(lambda: defaultdict(int))) to maintain the team distribution for each event.

Complexity

• Time complexity: \(O(N)\)
  • Reading and grouping player data takes \(O(N)\)
  • Processing each event is proportional to the number of events and teams, but totals \(O(N)\)
• Space complexity: \(O(N)\)
  • Memory needed to store and group player data

Implementation Notes

• By nesting defaultdict, we can efficiently perform two-dimensional aggregation by event number and team number

• For each event, we compute the number of cross-team pairs by subtracting the number of same-team pairs from the total number of pairs

• Be careful about integer overflow (in Python, integers have arbitrary precision, but the values can become very large)

  Source Code

import sys
from collections import defaultdict

def main():
    data = sys.stdin.read().split()
    n = int(data[0])
    
    players = []
    for i in range(n):
        p = int(data[1 + 2*i])
        q = int(data[2 + 2*i])
        players.append((p, q))
    
    comp_team_count = defaultdict(lambda: defaultdict(int))
    
    for p, q in players:
        comp_team_count[p][q] += 1
    
    total = 0
    for comp in comp_team_count:
        teams = comp_team_count[comp]
        total_players_in_comp = sum(teams.values())
        same_team_pairs = 0
        for count in teams.values():
            same_team_pairs += count * (count - 1) // 2
        
        total += total_players_in_comp * (total_players_in_comp - 1) // 2 - same_team_pairs
    
    print(total)

if __name__ == "__main__":
    main()

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