Overview

This problem asks you to select an optimal group from the given employees to maximize the minimum of the total programming ability and the total design ability. Since all employees’ ability values are positive integers, “selecting all employees for the team” is always optimal.

Analysis

At first glance, this might appear to be a complex problem requiring careful decisions about “which employees to select and which to exclude” (such as dynamic programming similar to the knapsack problem). However, by focusing on the constraints and properties of the problem, it can be solved very simply.

Key Insight: All ability values are “positive integers”

Looking at the constraints, each employee’s ability values satisfy \(1 \leq A_i, B_i \leq 10^9\), meaning they are all positive integers.

Let’s consider adding one new employee to the team. When employee \(i\) is added to a team, the team’s total programming ability increases by \(A_i\), and the total design ability increases by \(B_i\). Since all ability values are positive, the more members you add, the more both the total programming ability and the total design ability will necessarily increase.

Mathematical Explanation

Let \(S\) be the set of selected employees. If some employee \(j\) is not included in set \(S\), and we add this employee \(j\) to create a new set \(S' = S \cup \{j\}\), the following holds:

• \(\sum_{i \in S'} A_i = \left(\sum_{i \in S} A_i\right) + A_j > \sum_{i \in S} A_i\)
• \(\sum_{i \in S'} B_i = \left(\sum_{i \in S} B_i\right) + B_j > \sum_{i \in S} B_i\)

Since both the total programming ability and the total design ability strictly increase, their minimum — the “overall strength” — will never decrease; it will either stay the same or increase.

Therefore, to maximize the team’s overall strength, it is best not to exclude any employee and to select all \(N\) employees for the team.

Algorithm

1. Initialize a variable sum_A to hold the total programming ability and a variable sum_B to hold the total design ability, both set to \(0\).
2. For each employee \(i\), add the input values \(A_i\) and \(B_i\) to sum_A and sum_B respectively.
3. After summing all employees’ ability values, compute and output min(sum_A, sum_B).

With this approach, there is no need to store all elements in an array; you can obtain the answer simply by accumulating the sums as you read the input.

Complexity

• Time Complexity: \(O(N)\) Since we only scan through the data of \(N\) employees once and accumulate the sums, the processing completes in linear time with respect to \(N\). With \(N \leq 2 \times 10^5\), this easily fits within the time limit.

• Space Complexity: \(O(1)\) Since we process the input on the fly and only maintain the totals, almost no additional memory is consumed.

Implementation Notes

• Preventing Overflow (Attention to Data Types) The maximum value of \(N\) is \(2 \times 10^5\), and the maximum value of each ability is \(10^9\). The total sum of all employees’ abilities can reach up to \(2 \times 10^5 \times 10^9 = 2 \times 10^{14}\). This value greatly exceeds the maximum of a typical 32-bit integer type (such as int in C++, which holds up to approximately \(2 \times 10^9\)), causing overflow. Therefore, the variables sum_A and sum_B that store the totals must use a 64-bit integer type (long long in C++).

  Source Code

#include <iostream>
#include <algorithm>

using namespace std;

int main() {
    // Optimize standard I/O operations for competitive programming
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    int N;
    if (!(cin >> N)) return 0;

    long long sum_A = 0;
    long long sum_B = 0;

    for (int i = 0; i < N; ++i) {
        long long a, b;
        cin >> a >> b;
        sum_A += a;
        sum_B += b;
    }

    cout << min(sum_A, sum_B) << "\n";

    return 0;
}

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This editorial was generated by gemini-3.5-flash-thinking.