Print the lexicographically smallest DFS of the graph starting from 1 in C Program.

CServer Side ProgrammingProgramming

We will be given a connected graph with N vertices and M edges. So we have to print the lexicographically smallest DFS of the graph starting from 1.

Vertices should be numbered from 1 to N

Example

Input: N = 5 M =5
   edge(1, 4, arr)
   edge(3, 4, arr)
   edge(5, 4, arr)
   edge(3, 2, arr)
   edge(1, 5, arr)
   edge(1, 2, arr)
   edge(3, 5, arr)
   edge(1, 3, arr)
output: 1 2 3 4 5

Instead of doing a normal DFS, first we will sort the edges associated with each vertex, so that in each turn only the smallest edge is picked first. After sorting, just perform a normal DFS which will give the lexicographically smallest DFS traversal.

Given below is the C++ implementation of the algorithm given below.

Algorithm

Start
Step 1 -> Declare Function void lexo(vector<int>* arr, int n)
   Declare bool check[n + 1] = { 0 }
   Loop For int i=0 and i<n and i++
      Call sort(arr[i].begin(), arr[i].end())
      Loop For int i = 1 and i < n and i++
         IF !check[i]
            Call graph(arr, i, n, check)
      End
   End
Step 2 -> declare Function void edge(int u, int v, vector<int>* arr)
   Call ar[u].push_back(v)
   Call ar[v].push_back(u)
Step 3 -> Declare function void graph(vector<int>* arr, int src, int n,bool* check) print src
   Set check[src] = true
   Loop for int i = 0 and i < arr[src].size() and i++
      IF !check[arr[src][i]]
         Call graph(arr, arr[src][i], n, check)
      End
   End
Step 4- > In main()
   Declare int n = 5, m = 5
   Use STL vector<int> arr[n + 1]
   Call edges(1,4, arr)
   Call edges(3,4, arr)....
   Call lexo(arr, n)
Stop

Example

#include <bits/stdc++.h>
using namespace std;
//for inserting an edge
void edge(int u, int v, vector<int>* arr){
   arr[u].push_back(v);
   arr[v].push_back(u);
}
// Function for dfs graph traversal
void graph(vector<int>* arr, int src, int n,bool* check){
   cout << src << " ";
   check[src] = true;
   for (int i = 0; i < arr[src].size(); i++){
      if (!check[arr[src][i]])
         graph(arr, arr[src][i], n, check);
   }
}
void lexo(vector<int>* arr, int n){
   bool check[n + 1] = { 0 };
   for (int i = 0; i < n; i++)
      sort(arr[i].begin(), arr[i].end());
   for (int i = 1; i < n; i++){
      if (!check[i])
      graph(arr, i, n, check);
   }
}
int main(){
   int n = 5, m = 5;
   vector<int> arr[n + 1];
   // for inserting an edge
   edge(1, 4, arr);
   edge(3, 4, arr);
   edge(5, 4, arr);
   edge(3, 2, arr);
   edge(1, 5, arr);
   edge(1, 2, arr);
   edge(3, 5, arr);
   edge(1, 3, arr);
   //call lexo function
   lexo(arr, n);
   return 0;
}

Output

if we run above program then it will generate following output

1 2 3 4 5
raja
Published on 22-Aug-2019 07:36:26
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