Hello. Today, we will tackle an algorithm problem related to helping the less fortunate through a coding testing course using Java. Helping the less fortunate refers to the activities that support neighbors in need. Through this problem, we will understand and implement the ‘assignment’ algorithm.
Problem Description
Three volunteers, Gasanho, Seonghoon, and Minjae, have decided to help the less fortunate at specified times. However, since each volunteer cannot help all the less fortunate alone, they must efficiently allocate their time.
Considering the following conditions, implement an algorithm that assigns volunteers to provide the most help to the less fortunate:
- Number of volunteers: N (1 ≤ N ≤ 100)
- Number of less fortunate: M (1 ≤ M ≤ 100)
- Each volunteer can help a maximum of K less fortunate individuals.
- An adjacency matrix representing the connection between volunteers and the less fortunate will be provided.
Input Format
The first line contains the number of volunteers N and the number of less fortunate M. The next N lines provide information about which less fortunate individuals each volunteer can help. A value of 1 indicates they can help, while 0 indicates they cannot.
Output Format
Print the number of volunteers who helped the most less fortunate individuals and the respective indices of the individuals they can assist.
Example Input
3 4
1 1 0 1
1 0 1 1
0 1 1 0
Example Output
2
1 2
2 3
Problem Solving Process
To solve this problem, we can use search techniques such as DFS (Depth-First Search) or BFS (Breadth-First Search). Follow the steps below to design an algorithm to solve the assignment problem:
Step 1: Process Input Data
Scanner sc = new Scanner(System.in);
int N = sc.nextInt(); // Number of volunteers
int M = sc.nextInt(); // Number of less fortunate
int[][] graph = new int[N][M]; // Create adjacency matrix
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++) {
graph[i][j] = sc.nextInt(); // Input adjacency matrix
}
}
Step 2: Assign Volunteers
Based on the situation where volunteers can help each less fortunate individual, use DFS to explore all possible combinations:
boolean[] visited = new boolean[M]; // Check visit status
int[] result = new int[N]; // Result array
int[] assignment = new int[N]; // Volunteer assignment
for (int i = 0; i < N; i++) {
Arrays.fill(visited, false); // Initialize visit array
assignVolunteerToNeighbour(i, graph, visited, assignment);
}
private static boolean assignVolunteerToNeighbour(int volunteer, int[][] graph, boolean[] visited, int[] assignment) {
for (int neighbour = 0; neighbour < M; neighbour++) {
if (graph[volunteer][neighbour] == 1 && !visited[neighbour]) {
visited[neighbour] = true;
if (assignment[neighbour] == -1 || assignVolunteerToNeighbour(assignment[neighbour], graph, visited, assignment)) {
assignment[neighbour] = volunteer;
return true;
}
}
}
return false;
}
Step 3: Print Final Result
for (int i = 0; i < N; i++) {
if (assignment[i] != -1) {
System.out.println((assignment[i] + 1) + " " + (i + 1));
}
}
Conclusion
Today, we implemented an algorithm to assign volunteers through a Java coding test problem related to helping the less fortunate. I hope this problem has helped deepen your understanding of graph theory and search algorithms. Additionally, I hope it has assisted you in practicing skills that can be directly applied in coding tests and work. I will return with a new algorithm problem in the next course. Thank you!