Planning a trip can be an exciting yet challenging task for many people. Various elements such as destination selection, itinerary, and budget management must be considered. In this course, we will explore how to solve these issues through programming.
Problem Definition
Please create an algorithm to plan a trip that satisfies the following conditions.
- A list of destinations is provided.
- Each destination has a specific cost and time required.
- You want to visit as many destinations as possible without exceeding a maximum travel budget (B) within a fixed duration (C).
- Output the combinations of destinations that can be visited.
Input and Output Format
Input:
Number of destinations N (1 ≤ N ≤ 20)
Cost and time required for each destination (cost, time)
Maximum travel budget B (1 ≤ B ≤ 1000)
Maximum time limit C (1 ≤ C ≤ 100)
Output:
List of possible maximum destinations
Example
Input:
4
100 1
200 2
150 1
300 4
400
5
Output:
Destinations: (100, 1), (150, 1), (200, 2)
Problem-Solving Approach
This problem can be classified as a combination problem since it requires finding possible combinations with given resources (cost, time).
To solve this, we will use the Backtracking technique. Backtracking is a method that explores all possible cases to find a solution. For each destination, we will have two choices:
- Visit the destination.
- Do not visit the destination.
We will check the cost and time at each path to ensure we do not exceed the maximum limits. This process will be repeated recursively to explore all combinations.
Swift Code Implementation
Now let’s implement the above algorithm in Swift.
struct Destination {
let cost: Int
let time: Int
}
func planTrip(destinations: [Destination], budget: Int, timeLimit: Int) -> [(Int, Int)] {
var maxDestinations: [(Int, Int)] = []
func backtrack(index: Int, currentBudget: Int, currentTime: Int, currentList: [(Int, Int)]) {
if currentBudget > budget || currentTime > timeLimit {
return
}
if currentList.count > maxDestinations.count {
maxDestinations = currentList
}
for i in index..<destinations.count {="" backtrack(index:="" i="" +="" 1,="" currentbudget:="" currentbudget="" destinations[i].cost,="" currenttime:="" currenttime="" destinations[i].time,="" currentlist:="" currentlist="" [(destinations[i].cost,="" destinations[i].time)])="" }="" 0,="" [])="" print("destinations:",="" maxdestinations)="" return="" maxdestinations="" example="" execution="" let="" result="planTrip(destinations:" [destination(cost:="" 100,="" time:="" 1),="" destination(cost:="" 200,="" 2),="" 150,="" 300,="" 4)],="" budget:="" 400,="" timelimit:="" 5)="" <="" code=""></destinations.count>
<h2>Code Explanation</h2>
<p>The implemented code performs the following roles:</p>
<ol>
<li><strong>Struct Definition:</strong> The <code>Destination</code> struct represents the cost and time required for a destination.</li>
<li><strong>Main Function Definition:</strong> The <code>planTrip</code> function finds and returns the maximum possible destinations based on the given budget and time.</li>
<li><strong>Backtracking Implementation:</strong> An internal function <code>backtrack</code> is defined to explore all cases.</li>
</ol>
<h3>Result Analysis</h3>
<p>The outputted result represents the possible destinations within the maximum budget and time. This method can be usefully employed in trip planning and presents the optimal combinations of destinations that a traveler can visit.</p>
<h2>Conclusion</h2>
<p>Now you have created an algorithm that allows you to consider various destinations and formulate an optimal plan based on the given input. Utilize this technique, useful for coding tests and real-world problem-solving, to make your trip planning even more effective!</p>
<div class="tip">
<strong>Tip:</strong> When planning your actual trip, considering various factors such as destination ratings, distance, climate, and more, in addition to cost, will lead to a more satisfying plan.
</div>
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