INLO23 - Editorial

bfs
editorial
graph-theory
inlo2016
inlo23
medium-hard

#1

PROBLEM LINK:

Practice
Contest

Author: RAVIT SINGH MALIK
Editorialist: RAVIT SINGH MALIK

DIFFICULTY:

MEDIUM-HARD

PREREQUISITES:

BFS , GRAPHS

PROBLEM:

F(x,y) is a function which determines the performance of the super-computer.
Here (x,y) is that pair of processes (y is always a sub-process of process x) where the difference between the factors of process x and process y is maximum.
You are required to find maximum value of G(x)-G(Y) .

EXPLANATION:

This problem is completely based on Breadth-first search [BFS], which is used to traversing or searching tree or graph data structures. It starts at the tree root
(or some arbitrary node of a graph, sometimes referred to as a ‘search key’)
and explores the neighbor nodes first, before moving to the next level neighbors.

Pseudocode :
Input: A graph Graph and a starting vertex root of Graph.

Output: All vertices reachable from root labeled as explored.

A non-recursive implementation of breadth-first search:


 1.   Breadth-First-Search(Graph, root):  
 2.     
 3.     for each node n in Graph:              
 4.         n.distance = INFINITY          
 5.         n.parent = NIL  
 6. 
 7.     create empty queue Q      
 8. 
 9.     root.distance = 0
10.     Q.enqueue(root)                      
11. 
12.     while Q is not empty:        
13.     
14.         current = Q.dequeue()
15.     
16.         for each node n that is adjacent to current:
17.             if n.distance == INFINITY:
18.                 n.distance = current.distance + 1
19.                 n.parent = current
20.                 Q.e
 

Time and space complexity.

The time complexity can be expressed as O(|V|+|E|), since every vertex and every edge will be explored in the worst case.
|V| is the number of vertices and |E| is the number of edges in the graph. Note that O(|E|) may vary between O(1) and O(|V|^{2}), depending on how sparse the input graph is.

So,the given question is an improvisation of Breadth-first search [BFS], you have to just store the maximum difference
of parent to its every childern. and store the maximum value between the parent and child at the child node.

Pseudocode :

while(!q.empty()) int s=q.front() q.pop() int size=adj[s].size() for( i=0 ;i < size ; i++) int num = adj[s]* if(visit[num]!=true) visit[num]=true if(maxi < a[s] - a[num]) maxi = a[s] - a[num] if(a[s] > a[num]) a[num]=a[s] qu.push(adj[s]*) =>> $maxi$ is the required answer.

AUTHOR’S AND TESTER’S SOLUTIONS:

Author’s solution can be found here.

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