Author: root

Introduction to the Floyd-Warshall Algorithm

The Floyd-Warshall Algorithm is an algorithm used to find the shortest paths between all pairs of vertices in a graph.
This algorithm is suitable for calculating the shortest distance for all pairs of a given graph,
and it works even in the presence of negative weights.
Its complexity is O(V^3), where V represents the number of vertices.
The Floyd-Warshall Algorithm utilizes dynamic programming techniques to compute the shortest paths,
incrementally building up the optimal solution.

Problem Description

Given the paths and distances between certain cities,
we will solve the problem of finding the shortest distance between all cities.
This can be very useful for finding optimal routes in heavily trafficked cities.

Problem

There are N cities, and given the distances between the cities,
write a program to output the shortest distance between all cities.
An example input is as follows:

3
0 4 2
float(inf) 0 5
float(inf) float(inf) 0
        

An example output is as follows:

0 4 2
float(inf) 0 5
float(inf) float(inf) 0
        

Solution Process

1. Input Processing

We process the input to solve the problem.
The first line gives the number of cities N, and the following N lines provide the distance information between each city.
‘float(inf)’ indicates there is no path between the two cities.

2. Initial Matrix Creation

Based on the given distance information, we create a 2D array to form the initial matrix.
This matrix contains distance information from city A to city B,
with initial values set to the given distances, and ‘float(inf)’ for cases where there is no path.

3. Applying the Floyd-Warshall Algorithm

The next step is to apply the Floyd-Warshall Algorithm.
The algorithm considers each city K as an intermediate vertex,
determining whether the path from city I to J is shorter by going through city K (i.e.,
if distance[I][J] > distance[I][K] + distance[K][J]),
and updates to the shorter path accordingly.

4. Outputting the Result

We output the shortest distance matrix for all cities.
The result is formatted to distinguish between each city’s shortest distances and ‘float(inf)’.

5. Kotlin Implementation

fun floydWarshall(graph: Array) {
    val n = graph.size
    for (k in 0 until n) {
        for (i in 0 until n) {
            for (j in 0 until n) {
                if (graph[i][j] > graph[i][k] + graph[k][j]) {
                    graph[i][j] = graph[i][k] + graph[k][j]
                }
            }
        }
    }
}

// Main function to execute the program
fun main() {
    val n = readLine()!!.toInt()
    val graph = Array(n) { IntArray(n) { Int.MAX_VALUE } }
    
    // Initialize the graph with input
    for (i in 0 until n) {
        val distances = readLine()!!.split(" ").map { it.toInt() }
        for (j in 0 until n) {
            graph[i][j] = distances[j]
        }
        graph[i][i] = 0 // Distance from a city to itself is always 0
    }
    
    floydWarshall(graph)
    
    // Output the final distance matrix
    for (i in 0 until n) {
        for (j in 0 until n) {
            if (graph[i][j] == Int.MAX_VALUE) {
                print("float(inf) ")
            } else {
                print("${graph[i][j]} ")
            }
        }
        println()
    }
}
        

Introduction

Hello, in this Kotlin coding test tutorial, we will cover the algorithmic problem of ‘Finding Cities at a Specific Distance’.
This problem involves graph traversal and requires an algorithmic approach to find nodes in cities at a specific distance based on given conditions.
We will look at how to solve the problem step by step using Kotlin.

Problem Description

The given graph consists of N cities and M bidirectional roads.
Each city is identified by numbers from 1 to N, and the roads connect two cities.
The goal is to find all cities that are exactly at the given distance K.

Input Format
The first line contains the number of cities N (1 ≤ N ≤ 30000) and the number of roads M (1 ≤ M ≤ 200000), followed by the distance K (0 ≤ K ≤ 30000) we want to find.
From the second line, M lines follow, each containing information about two connected cities A and B (1 ≤ A, B ≤ N).

Output Format
Output the numbers of cities that are exactly at distance K in ascending order; if there are no such cities, output -1.

Problem Solving Approach

This problem involves exploring vertices in a graph that match the given conditions.
By using the BFS (Breadth-First Search) algorithm, we can find cities at a specific distance by searching to the appropriate depth.
BFS can be implemented using an empty queue and an adjacency list.

Implementation Method

First, we construct an adjacency list based on the given city and road information.
Then we use the BFS algorithm to find the cities that reach the given distance K.

The implementation steps are as follows:

1. Process input values: Read the number of cities, number of roads, and distance K.
2. Construct adjacency list: Store other cities connected to each city in list form.
3. Initialize BFS: Explore from the starting city to the distance K.
4. Store results: Save city numbers that have reached the distance K in a list.
5. Output results: Sort and print the cities reached at distance K.

Code Implementation

Below is the solution code implemented using the BFS algorithm in Kotlin:

import java.util.*

fun main() {
    val scanner = Scanner(System.`in`)
    val n = scanner.nextInt() // Number of cities
    val m = scanner.nextInt() // Number of roads
    val k = scanner.nextInt() // Desired distance
    val x = scanner.nextInt() // Starting city

    // Initialize adjacency list
    val graph = Array(n + 1) { mutableListOf() }

    // Input road information
    for (i in 0 until m) {
        val a = scanner.nextInt()
        val b = scanner.nextInt()
        graph[a].add(b)
        graph[b].add(a)
    }

    // BFS implementation
    val distance = IntArray(n + 1) { -1 }
    distance[x] = 0
    val queue: Queue = LinkedList()
    queue.add(x)

    while (queue.isNotEmpty()) {
        val current = queue.poll()
        for (neighbor in graph[current]) {
            if (distance[neighbor] == -1) {
                distance[neighbor] = distance[current] + 1
                queue.add(neighbor)
            }
        }
    }

    // Find result cities
    val result = mutableListOf()
    for (i in 1..n) {
        if (distance[i] == k) {
            result.add(i)
        }
    }

    // Output results
    if (result.isEmpty()) {
        println(-1)
    } else {
        result.sort()
        for (city in result) {
            println(city)
        }
    }
}
        

Solution Analysis

The above code is a typical BFS implementation designed to solve the given problem.
In particular, by representing the road connectivity as an adjacency list, it optimizes memory usage and allows for fast searching.
BFS operates by exploring all nodes and placing adjacent nodes into a queue.
This way, it updates the distance to each node and ultimately collects nodes that stop at distance K.

If no city is found at distance K, returning -1 is also a good programming practice.
This code structure can be easily modified to suit different types of graph traversal problems.

Conclusion

Thus, we have learned about the process of implementing Kotlin code using the BFS algorithm to solve the ‘Finding Cities at a Specific Distance’ problem.
This problem helps in understanding the basics of graph problems and can develop problem-solving skills for various problem types.
It will also be beneficial for effectively applying algorithms in work or personal projects.
We plan to cover more diverse algorithm problems in the future, so please stay tuned.