Minimum time taken by each job to be completed given by a Directed Acyclic Graph

03 Jul, 2021
Given a Directed Acyclic Graph having V vertices and E edges, where each edge {U, V} represents the Jobs U and V such that Job V can only be started only after completion of Job U. The task is to determine the minimum time taken by each job to be completed where each Job takes unit time to get completed.
Examples:
Input: N = 10, E = 13, Below is the given graph:
Output: 1 1 2 2 2 3 4 5 2 6  Explanation:Start the jobs 1 and 2 at the beginning and complete them at 1 unit of time. Since, jobs 3, 4, 5, and 9 have the only dependency on one job (i.e 1st job for jobs 3, 4, and 5 and 2nd job for job 9). So, we can start these jobs at 1st unit of time and complete these at 2nd unit of time after the completion of the dependent Job.Similarly, Job 6 can only be done after 3rd and 4th jobs are done. So, start it at 2nd unit of time and complete it at 3rd unit of time.Job 7 can only be done after job 6 is done. So, you can start it at 3rd unit of time and complete it at 4th unit of time.Job 8 can only be done after 4th, 5th, and 7th jobs are done. So, start it at 4th unit of time and complete it at 5th unit of time.Job 10 can only be done after the 8th job is done. So, start it at 5th unit of time and complete it at 6th unit of time.
Input: N = 7, E = 7, Below is the given graph:
Output: 1 2 3 3 3 4 4  Explanation:Start the Job 1 at the beginning and complete it at 1st unit of time.The job 2 can only be done after 1st Job is done. So, start it at 1st unit of time and complete it at 2nd unit of time.Since, Job 3, 4, and 5 have the only dependency on 2nd Job. So, start these jobs at 2nd unit of time and complete these at 3rd unit of time.The Job 6 can only be done after the 3rd and 4th Job is done. So, start it at 3rd unit of time and complete it at 4th unit of time.The Job 7 can only be done after the 5th Job is done. So, start it at 3rd hour and complete it at 4th unit of time.
Approach: The job can be started only if all the jobs that are prerequisites of the job that are done. Therefore, the idea is to use Topological Sort for the given network. Below are the steps:
Finish the jobs that are not dependent on any other job.Create an array inDegree[] to store the count of the dependent node for each node in the given network.Initialize a queue and push all the vertex whose inDegree[] is 0.Initialize the timer to 1 and store the current queue size(say size) and do the following:Pop the node from the queue until the size is 0 and update the finishing time of this node to the timer.While popping the node(say node U) from the queue decrement the inDegree of every node connected to it.If inDegree of any node is 0 in the above step then insert that node in the queue.Increment the timer after all the above steps.Print the finishing time of all the nodes after we traverse every node in the above step.
Finish the jobs that are not dependent on any other job.
Create an array inDegree[] to store the count of the dependent node for each node in the given network.
Initialize a queue and push all the vertex whose inDegree[] is 0.
Initialize the timer to 1 and store the current queue size(say size) and do the following:Pop the node from the queue until the size is 0 and update the finishing time of this node to the timer.While popping the node(say node U) from the queue decrement the inDegree of every node connected to it.If inDegree of any node is 0 in the above step then insert that node in the queue.Increment the timer after all the above steps.
Pop the node from the queue until the size is 0 and update the finishing time of this node to the timer.
While popping the node(say node U) from the queue decrement the inDegree of every node connected to it.
If inDegree of any node is 0 in the above step then insert that node in the queue.
Increment the timer after all the above steps.
Print the finishing time of all the nodes after we traverse every node in the above step.
Below is the implementation of the above approach:
C++
Java
Python3
C#
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;#define maxN 100000 // Adjacency List to store the graphvector<int> graph[maxN]; // Array to store the in-degree of nodeint indegree[maxN]; // Array to store the time in which// the job i can be doneint job[maxN]; // Function to add directed edge// between two verticesvoid addEdge(int u, int v){    // Insert edge from u to v    graph[u].push_back(v);     // Increasing the indegree    // of vertex v    indegree[v]++;} // Function to find the minimum time// needed by each node to get the taskvoid printOrder(int n, int m){    // Find the topo sort order    // using the indegree approach     // Queue to store the    // nodes while processing    queue<int> q;     // Pushing all the vertex in the    // queue whose in-degree is 0     // Update the time of the jobs    // who don't require any job to    // be completed before this job    for (int i = 1; i <= n; i++) {        if (indegree[i] == 0) {            q.push(i);            job[i] = 1;        }    }     // Iterate until queue is empty    while (!q.empty()) {         // Get front element of queue        int cur = q.front();         // Pop the front element        q.pop();         for (int adj : graph[cur]) {             // Decrease in-degree of            // the current node            indegree[adj]--;             // Push its adjacent elements            if (indegree[adj] == 0) {                job[adj] = job[cur] + 1;                q.push(adj);            }        }    }     // Print the time to complete    // the job    for (int i = 1; i <= n; i++)        cout << job[i] << " ";    cout << "\n";} // Driver Codeint main(){    // Given Nodes N and edges M    int n, m;    n = 10;    m = 13;     // Given Directed Edges of graph    addEdge(1, 3);    addEdge(1, 4);    addEdge(1, 5);    addEdge(2, 3);    addEdge(2, 8);    addEdge(2, 9);    addEdge(3, 6);    addEdge(4, 6);    addEdge(4, 8);    addEdge(5, 8);    addEdge(6, 7);    addEdge(7, 8);    addEdge(8, 10);     // Function Call    printOrder(n, m);    return 0;}
// Java program for the above approachimport java.util.*; class GFG{     static final int maxN = 100000; // Adjacency List to store the graph@SuppressWarnings("unchecked")static Vector<Integer> []graph = new Vector[maxN]; // Array to store the in-degree of nodestatic int []indegree = new int[maxN]; // Array to store the time in which// the job i can be donestatic int []job = new int[maxN]; // Function to add directed edge// between two verticesstatic void addEdge(int u, int v){         // Insert edge from u to v    graph[u].add(v);     // Increasing the indegree    // of vertex v    indegree[v]++;} // Function to find the minimum time// needed by each node to get the taskstatic void printOrder(int n, int m){         // Find the topo sort order    // using the indegree approach     // Queue to store the    // nodes while processing    Queue<Integer> q = new LinkedList<>();         // Pushing all the vertex in the    // queue whose in-degree is 0     // Update the time of the jobs    // who don't require any job to    // be completed before this job    for(int i = 1; i <= n; i++)    {        if (indegree[i] == 0)        {            q.add(i);            job[i] = 1;        }    }     // Iterate until queue is empty    while (!q.isEmpty())    {         // Get front element of queue        int cur = q.peek();         // Pop the front element        q.remove();         for(int adj : graph[cur])        {                         // Decrease in-degree of            // the current node            indegree[adj]--;             // Push its adjacent elements            if (indegree[adj] == 0){                job[adj] = 1 + job[cur];                q.add(adj);            }        }    }     // Print the time to complete    // the job    for(int i = 1; i <= n; i++)        System.out.print(job[i] + " ");    System.out.print("\n");} // Driver Codepublic static void main(String[] args){         // Given Nodes N and edges M    int n, m;    n = 10;    m = 13;         for(int i = 0; i < graph.length; i++)        graph[i] = new Vector<Integer>();             // Given directed edges of graph    addEdge(1, 3);    addEdge(1, 4);    addEdge(1, 5);    addEdge(2, 3);    addEdge(2, 8);    addEdge(2, 9);    addEdge(3, 6);    addEdge(4, 6);    addEdge(4, 8);    addEdge(5, 8);    addEdge(6, 7);    addEdge(7, 8);    addEdge(8, 10);     // Function call    printOrder(n, m);}} // This code is contributed by Amit Katiyar
# Python3 program for the above approachfrom collections import defaultdict # Class to represent a graphclass Graph:         def __init__(self, vertices, edges):                 # Dictionary containing adjacency List        self.graph = defaultdict(list)                 # No. of vertices        self.n = vertices                  # No. of edges        self.m = edges              # Function to add an edge to graph    def addEdge(self, u, v):        self.graph[u].append(v)         # Function to find the minimum time    # needed by each node to get the task    def printOrder(self, n, m):               # Create a vector to store indegrees of all        # vertices. Initialize all indegrees as 0.        indegree = [0] * (self.n + 1)                 # Traverse adjacency lists to fill indegrees        # of vertices. This step takes O(V + E) time        for i in self.graph:            for j in self.graph[i]:                indegree[j] += 1                         # Array to store the time in which        # the job i can be done        job = [0] * (self.n + 1)                 # Create an queue and enqueue all        # vertices with indegree 0        q = []                 # Update the time of the jobs        # who don't require any job to        # be completed before this job        for i in range(1, self.n + 1):            if indegree[i] == 0:                q.append(i)                job[i] = 1                         # Iterate until queue is empty        while q:                         # Get front element of queue            cur = q.pop(0)                         for adj in self.graph[cur]:                                 # Decrease in-degree of                # the current node                indegree[adj] -= 1                               # Push its adjacent elements                if (indegree[adj] == 0):                    job[adj] =  1 + job[cur]                    q.append(adj)                             # Print the time to complete        # the job        for i in range(1, n + 1):            print(job[i], end = " ")                     print() # Driver Code # Given Nodes N and edges Mn = 10m = 13 g = Graph(n, m) # Given Directed Edges of graphg.addEdge(1, 3)g.addEdge(1, 4)g.addEdge(1, 5)g.addEdge(2, 3)g.addEdge(2, 8)g.addEdge(2, 9)g.addEdge(3, 6)g.addEdge(4, 6)g.addEdge(4, 8)g.addEdge(5, 8)g.addEdge(6, 7)g.addEdge(7, 8)g.addEdge(8, 10) # Function Callg.printOrder(n, m) # This code is contributed by Aanchal Tiwari
// C# program for the above approachusing System;using System.Collections.Generic; class GFG{     static readonly int maxN = 100000; // Adjacency List to store the graphstatic List<int> []graph = new List<int>[maxN]; // Array to store the in-degree of nodestatic int []indegree = new int[maxN]; // Array to store the time in which// the job i can be donestatic int []job = new int[maxN]; // Function to add directed edge// between two verticesstatic void addEdge(int u, int v){         // Insert edge from u to v    graph[u].Add(v);     // Increasing the indegree    // of vertex v    indegree[v]++;} // Function to find the minimum time// needed by each node to get the taskstatic void printOrder(int n, int m){         // Find the topo sort order    // using the indegree approach     // Queue to store the    // nodes while processing    Queue<int> q = new Queue<int>();         // Pushing all the vertex in the    // queue whose in-degree is 0     // Update the time of the jobs    // who don't require any job to    // be completed before this job    for(int i = 1; i <= n; i++)    {        if (indegree[i] == 0)        {            q.Enqueue(i);            job[i] = 1;        }    }     // Iterate until queue is empty    while (q.Count != 0)    {         // Get front element of queue        int cur = q.Peek();         // Pop the front element        q.Dequeue();         foreach(int adj in graph[cur])        {                         // Decrease in-degree of            // the current node            indegree[adj]--;             // Push its adjacent elements            if (indegree[adj] == 0){                job[adj] = 1 + job[cur];                q.Enqueue(adj);            }        }    }     // Print the time to complete    // the job    for(int i = 1; i <= n; i++)        Console.Write(job[i] + " ");             Console.Write("\n");} // Driver Codepublic static void Main(String[] args){         // Given Nodes N and edges M    int n, m;    n = 10;    m = 13;         for(int i = 0; i < graph.Length; i++)        graph[i] = new List<int>();             // Given directed edges of graph    addEdge(1, 3);    addEdge(1, 4);    addEdge(1, 5);    addEdge(2, 3);    addEdge(2, 8);    addEdge(2, 9);    addEdge(3, 6);    addEdge(4, 6);    addEdge(4, 8);    addEdge(5, 8);    addEdge(6, 7);    addEdge(7, 8);    addEdge(8, 10);     // Function call    printOrder(n, m);}} // This code is contributed by Amit Katiyar
1 1 2 2 2 3 4 5 2 6
 
Time Complexity: O(V+E), where V is the number of nodes and E is the number of edges. Auxiliary Space: O(V)
amit143katiyar
aanchaltiwari
manvineekhra
ujjwal2001kant
Algorithms-Graph Traversals
BFS
Data Structures-Graph
Graph Traversals
Topological Sorting
Graph
Greedy
Greedy
Graph
BFS
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