Optimal Itinerary for Maximum Profit

Hard

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Problem statement

A traveling salesman plans a business trip of exactly k legs (or trips). He starts at his home base, city 0. For any trip from a city i to a city j, there are associated costs and earnings.

You are given three N x N 2D arrays:

Distance[i][j]: The cost of fuel to travel from city i to city j. Loss[i][j]: An entry fee or tax paid upon arriving at city j from city i. Gain[i][j]: The revenue earned from business conducted in city j after arriving from city i.

The net profit for a single trip from city i to city j is calculated as Gain[i][j] - Loss[i][j] - Distance[i][j].

Your task is to find the maximum possible total profit the salesman can achieve after completing exactly k trips. The salesman can revisit cities, including his home base.

Return the minimum possible value of x.

Detailed explanation ( Input/output format, Notes, Images )

Input Format:

The first line contains an integer N, the total number of cities (cities are numbered 0 to N-1).
The second line contains an integer k, the exact number of trips to be made.
The next N lines contain N space-separated integers each, representing the Distance matrix.
The next N lines contain N space-separated integers each, representing the Loss matrix.
The next N lines contain N space-separated integers each, representing the Gain matrix.

Output Format:

The output should be a single long integer representing the maximum possible profit.

Note:

The salesman always starts at city 0 before the first trip.
A trip from city i to city j is a single leg of the journey. A plan with k trips will have a path like c_0 -> c_1 -> ... -> c_k, where c_0 is always city 0.
The total profit is the sum of the net profits of all k individual trips.
If all possible itineraries result in a loss, the answer should be the maximum profit (i.e., the smallest loss, which will be a negative number).

Sample Input 1:

Sample Output 1:

Explanation for Sample 1:

The salesman must make exactly k=2 trips, starting from city 0. We need to find the path that maximizes total profit.

Let's first calculate the net profit for each possible one-way trip using the formula Profit = Gain - Loss - Distance.
  Profit(0 -> 1) = 95 - 5 - 10 = 80
  Profit(0 -> 2) = 45 - 5 - 10 = 30
  Profit(1 -> 0) = 5 - 5 - 10 = -10
  Profit(1 -> 2) = 10 - 20 - 10 = -20
  Profit(2 -> 0) = 65 - 5 - 10 = 50
  Profit(2 -> 1) = 75 - 5 - 10 = 60
 (Profit for trips to the same city, e.g., 0->0, is 0)
Now, let's analyze the total profit for possible 2-trip paths:
Path 1: A Greedy Approach
A simple greedy strategy would be to choose the most profitable first trip.
  The best first trip is 0 -> 1, with a profit of 80.
  From city 1, the best second trip is to stay put (1 -> 1), with a profit of 0.
  The total profit for this greedy path (0 -> 1 -> 1) is 80 + 0 = 80.
Path 2: The Optimal Path
Consider the path that starts with the less profitable first trip.
  The first trip is 0 -> 2, with a profit of 30.
  From city 2, the best second trip is 2 -> 1, with a profit of 60.
  The total profit for this path (0 -> 2 -> 1) is 30 + 60 = 90.
The maximum possible profit is 90. This example shows that a simple greedy approach (picking the best immediate trip) does not work. The optimal solution required taking a less profitable first step (0 -> 2) to unlock a highly profitable second step (2 -> 1). This is why a dynamic programming approach, which considers all possibilities, is necessary.

Expected Time Complexity:

The expected time complexity is O(k * N^2), where N is the number of cities and k is the number of trips.

Constraints:

1 <= N <= 50
1 <= k <= 50
0 <= Distance[i][j], Loss[i][j], Gain[i][j] <= 1000

Approaches (1)

Approach 1

Dynamic Programming

Time Complexity

Space Complexity

(100% EXP penalty)

Optimal Itinerary for Maximum Profit

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