Overview

This is a problem on a one-dimensional number line where we need to move Takahashi and Aoki to their respective target positions in the minimum number of moves. Using a “swap” operation on two adjacent cells, we can either move one person or swap the positions of both, aiming to reach the goal.

Analysis

The key to this problem lies in “whether the left-right positional relationship between Takahashi and Aoki needs to be swapped.”

1. When the positional relationship does not change

If the initial positions satisfy \(A < B\) and the target positions satisfy \(C < D\) (or vice versa), the two can reach their target positions without ever passing each other. In this case, Takahashi simply moves from \(A \to C\) and Aoki from \(B \to D\) by the shortest distance, so the required number of operations is the sum of their respective travel distances. $\(\text{Number of operations} = |A - C| + |B - D|\)$

2. When the positional relationship is swapped

If the initial positions satisfy \(A < B\) but the target positions satisfy \(C > D\) (or vice versa), the two must “pass each other (swap)” at some point.

Consider the situation where the two swap at adjacent cells \(x\) and \(x+1\). 1. Takahashi moves to cell \(x\), and Aoki moves to cell \(x+1\). 2. Swap cells \(x\) and \(x+1\) (1 operation). 3. After swapping, Takahashi moves from \(x+1 \to C\), and Aoki moves from \(x \to D\).

In this case, assuming Takahashi was at position \(A < B\), the total number of operations is: $\(\text{Cost} = |x - A| + |(x+1) - B| + 1 + |C - (x+1)| + |D - x|\)\( This simplifies to: \)\(\text{Cost} = |x - A| + |x - (B-1)| + |x - (C-1)| + |x - D| + 1\)$

We need to find the integer \(x\) that minimizes this expression. This reduces to the problem of minimizing the sum of distances from 4 points \(\{A, B-1, C-1, D\}\) on the number line (a variation of the median problem).

Algorithm

1. Determine positional relationship: If \((A < B) == (C < D)\) is true, no swap is needed.
   • The answer is \(|A - C| + |B - D|\).
2. When a swap is needed:
   • Prepare the 4 values \(\{A, B-1, C-1, D\}\) (if \(A > B\), adjust appropriately so they become \(B+1, C+1\), or generalize the approach).
   • In code, since we use the “left position” as the reference when the swap occurs, sort the following 4 points as \(a_0 \leq a_1 \leq a_2 \leq a_3\):
     • The one that should be on the left side before the swap, and the one that should be on the left side after the swap, among \(A\) and \(C\).
     • Specifically, when \(A < B\): \(\{A, B-1, C-1, D\}\), and when \(A > B\): \(\{A-1, B, C, D-1\}\).
   • The minimum sum of distances from 4 points is \((a_3 + a_2) - (a_1 + a_0)\).
   • The answer is this value plus \(1\) for the swap operation itself.

Complexity

• Time complexity: \(O(1)\)
  • Only numerical comparisons, sorting (4 elements), and basic arithmetic operations are performed, so it finishes in constant time.
• Space complexity: \(O(1)\)
  • Only a few variables worth of memory are used.

Implementation Notes

• 64-bit integer type: Since coordinates can reach \(10^{18}\), you need to use long long type in C++ or the standard integer type in Python.

• Absolute value calculation: When computing distances, be careful to use the absolute value function abs() or subtract the smaller value from the larger one to avoid negative numbers.

• Choosing the 4 points: When a swap is needed, subtracting \(1\) from one coordinate (or adding \(1\) to the other) makes it mathematically easier to handle the “state of being on two adjacent cells.”

  Source Code

#include <iostream>
#include <algorithm>
#include <vector>

using namespace std;

/**
 * Problem Summary:
 * Two people, Takahashi (T) at A and Aoki (A) at B, want to move to C and D respectively.
 * The operation allows swapping contents of adjacent cells x and x+1.
 * This operation can move one person (if only one is present) or swap their positions 
 * (if both are present). Each operation costs 1.
 * 
 * Logic:
 * 1. If the relative order of Takahashi and Aoki remains the same ((A < B) == (C < D)):
 *    The minimum operations required is simply the sum of their individual distances:
 *    |A - C| + |B - D|. They can reach their targets without ever needing to pass 
 *    through each other's cell.
 * 
 * 2. If the relative order of Takahashi and Aoki must change ((A < B) != (C < D)):
 *    They must swap positions an odd number of times. To minimize operations, 
 *    they should swap exactly once.
 *    Suppose they swap at positions x and x+1.
 *    - To swap, they must first move to x and x+1.
 *    - After swapping, they move from x+1 and x to their respective targets.
 *    The total operations would be:
 *    (Steps for T to reach x and A to reach x+1) + 1 (swap) + (Steps for T from x+1 to C and A from x to D)
 *    = |x - A| + |(x+1) - B| + 1 + |C - (x+1)| + |D - x|
 *    = |x - A| + |x - (B-1)| + |x - (C-1)| + |x - D| + 1
 * 
 *    To minimize this, x should be the median of the set {A, B-1, C-1, D}.
 *    For four points a1 <= a2 <= a3 <= a4, the minimum value of sum(|x - ai|) 
 *    is (a4 + a3 - a2 - a1).
 *    Thus, the minimum operations is (a4 + a3 - a2 - a1) + 1.
 */

typedef long long ll;

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

    ll A, B, C, D;
    if (!(cin >> A >> B >> C >> D)) return 0;

    // Check if the relative order of the two people is the same at the start and end
    if ((A < B) == (C < D)) {
        // Case 1: No relative order change needed
        ll diffAC = (A > C ? A - C : C - A);
        ll diffBD = (B > D ? B - D : D - B);
        cout << diffAC + diffBD << endl;
    } else {
        // Case 2: Relative order must change (exactly one swap is optimal)
        ll a[4];
        // The points are the start/end positions adjusted for the swap adjacency
        a[0] = (A < B ? A : B);
        a[1] = (A < B ? B : A) - 1;
        a[2] = (C < D ? C : D);
        a[3] = (C < D ? D : C) - 1;
        
        // Sort the four points to find the median range [a[1], a[2]]
        sort(a, a + 4);
        
        // The minimum sum of distances to four points a1, a2, a3, a4 is a4 + a3 - a2 - a1
        ll min_dist_sum = a[3] + a[2] - a[1] - a[0];
        cout << min_dist_sum + 1 << endl;
    }

    return 0;
}

This editorial was generated by gemini-3-flash-thinking.