Overview

This problem asks us to find the number of pairs of players who are entered in the same sport but belong to different teams. It can be solved efficiently using a complementary counting approach: subtract the “number of pairs in the same sport AND same team” from the “number of pairs in the same sport.”

Analysis

Problem with the Naive Approach

The simplest method is to enumerate all \(\frac{N(N-1)}{2}\) pairs from the \(N\) players and check for each pair whether they are “in the same sport” and “on different teams.” However, since \(N\) can be up to \(2 \times 10^5\), the number of pairs can reach approximately \(2 \times 10^{10}\), resulting in TLE (Time Limit Exceeded).

Key Insight: Complementary Counting

The condition for a valid match is “same sport AND different team.” Since counting this directly is difficult, we decompose it as follows:

\[\text{Number of valid match pairs} = \text{Number of same-sport pairs} - \text{Number of same-sport AND same-team pairs}\]

In other words, we first count all pairs belonging to the same sport, then subtract the pairs on the same team (which cannot compete against each other).

Concrete Example

For example, suppose sport 1 has 5 players. The number of pairs choosing 2 from these 5 is \(\binom{5}{2} = 10\).

Among them, if sport 1 has 3 players from team A and 2 players from team B, the number of same-team pairs is \(\binom{3}{2} + \binom{2}{2} = 3 + 1 = 4\).

Therefore, the number of valid match pairs in sport 1 is \(10 - 4 = 6\).

Algorithm

1. Count the number of players per sport: Store in sport_count[p] the number of players entered in sport \(p\).
2. Count the number of players per (sport, team) pair: Store in sport_team_count[(p, q)] the number of players belonging to sport \(p\) and team \(q\).
3. Sum up same-sport pairs: For each sport, compute \(\binom{n}{2} = \frac{n(n-1)}{2}\) and add to the total.
4. Subtract same-sport AND same-team pairs: For each (sport, team) combination, compute \(\binom{c}{2} = \frac{c(c-1)}{2}\) and subtract from the total.

The final value is the answer.

Complexity

• Time complexity: \(O(N)\) — We process each player once and iterate through each entry of the Counter once.
• Space complexity: \(O(N)\) — The number of entries stored in the Counter is at most \(N\).

Implementation Notes

• Using Counter (dictionary): By using Python’s collections.Counter, we can concisely aggregate counts per sport and per (sport, team) combination.

• Combination formula: The number of ways to choose 2 from \(n\) people, \(\binom{n}{2} = \frac{n(n-1)}{2}\), is a frequently used formula. It can be computed exactly using integer division //.

• Subtraction-based approach: When it is difficult to directly count “condition A AND condition B,” the technique of subtracting “condition A AND NOT condition B” from “all of condition A” is very commonly used in competitive programming.

  Source Code

from collections import Counter

def main():
    import sys
    input = sys.stdin.readline
    
    N = int(input())
    players = []
    for _ in range(N):
        p, q = map(int, input().split())
        players.append((p, q))
    
    # Count players per sport
    sport_count = Counter()
    # Count players per (sport, team) pair
    sport_team_count = Counter()
    
    for p, q in players:
        sport_count[p] += 1
        sport_team_count[(p, q)] += 1
    
    # For each sport, total pairs = C(n, 2)
    # Same-team pairs in that sport = sum of C(c, 2) for each team
    # Valid pairs = total pairs - same-team pairs
    
    answer = 0
    for sport, n in sport_count.items():
        answer += n * (n - 1) // 2
    
    for (sport, team), c in sport_team_count.items():
        answer -= c * (c - 1) // 2
    
    print(answer)

main()

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