Sign up for Kotlin coding test course, planning a trip

Problem Description

When planning a trip, we need to consider the time and cost required to travel between several cities. Let’s solve the problem of creating an optimal travel plan based on the information provided below.

Problem: Travel Route Optimization

You are given a graph representing N cities and the travel time/cost between each pair of cities. Write a program to calculate the optimal route that starts from the initial city, visits all cities in minimum time or cost, and returns to the starting city.

Input

• The first line contains the number of cities N (1 ≤ N ≤ 12).
• Next, an N x N matrix is provided where each element represents the travel time (or cost) between two cities. If travel is not possible, it is represented by -1.

Output

Print the minimum travel time (or cost).

Example

Problem Solving Process

1. Understanding the Problem

This problem involves finding a path that minimizes the travel time between the given cities. This is commonly known as the Traveling Salesman Problem (TSP), which is an NP-hard problem. The input graph is represented as an adjacency matrix, and we need to explore the optimal route based on this matrix.

2. Approach

To solve this problem, we will combine Depth-First Search (DFS) with a Dynamic Programming (DP) technique using bit masking. Bit masking allows us to efficiently represent the state of visited cities and helps avoid redundant calculations.

3. Kotlin Implementation

        
        fun findMinCost(graph: Array, visited: Int, pos: Int, n: Int, memo: Array): Int {
            // If all cities have been visited
            if (visited == (1 shl n) - 1) {
                return graph[pos][0] // Return to the starting city
            }

            // Memoization check
            if (memo[visited][pos] != -1) {
                return memo[visited][pos]
            }

            var ans = Int.MAX_VALUE

            // Explore all cities
            for (city in 0 until n) {
                // If not visited and travel is possible
                if ((visited and (1 shl city)) == 0 && graph[pos][city] != -1) {
                    val newCost = graph[pos][city] + findMinCost(graph, visited or (1 shl city), city, n, memo)
                    ans = Math.min(ans, newCost)
                }
            }

            // Save in memoization
            memo[visited][pos] = ans
            return ans
        }

        fun tsp(graph: Array, n: Int): Int {
            // Initialization
            val memo = Array(1 shl n) { IntArray(n) { -1 } }
            return findMinCost(graph, 1, 0, n, memo)
        }

        fun main() {
            val graph = arrayOf(
                intArrayOf(0, 10, 15, 20),
                intArrayOf(10, 0, -1, 25),
                intArrayOf(15, -1, 0, 30),
                intArrayOf(20, 25, 30, 0)
            )
            val n = graph.size
            val result = tsp(graph, n)
            println("Minimum travel cost: $result")
        }
        
    

4. Code Explanation

The above code shows the overall flow of the algorithm for optimizing the travel route. The main parts are as follows:

• findMinCost: Takes the current city and the state of visited cities as parameters and recursively calculates the minimum cost.
• tsp: Calculates the overall minimum cost using bit masking and memoization.

5. Performance Analysis

This code has a time complexity of O(N^2 * 2^N), where N is the number of cities. This is due to the nature of searching all paths with DFS and the feature of storing visited states using bit masking. It is practically executable for inputs with up to 12 cities, making it efficient for reasonably sized inputs.

Conclusion

Through the travel planning algorithm problem, we have learned how to utilize Dynamic Programming techniques with DFS and bit masking. This is a valuable skill for coding tests and algorithm challenges, so be sure to master it. Keep improving your algorithm skills through various problems!