C++ program to count maximum possible division can be made in a graph with given condition

Suppose we have an adjacency matrix of a graph G. We have to check whether we can divide the vertices into non-empty sets V1, ... Vk, such that: every edge connects two vertices belonging to two adjacent sets. If the answer is yes, we have to find the maximum possible value of sets k, in such division.

So, if the input is like

then the output will be 4

Steps

To solve this, we will follow these steps −

Define an array dp of size: 210.
n := size of matrix
fl := 1
ans := 0
for initialize i := 0, when i < n and fl is non-zero, update (increase i by 1), do:
   fill dp with -1
   dp[i] := 0
   Define one queue q
   insert i into q
   while (q is not empty), do:
      x := first element of q
      delete element from q
      for initialize j := 0, when j < n, update (increase j by 1), do:
         if matrix[x, j] is same as 1, then:
            if dp[j] is same as -1, then:
               dp[j] := dp[x] + 1
               insert j into q
            otherwise when |dp[j] - dp[x]| is not equal to 1, then:
               fl := 0
      for initialize j := 0, when j < n, update (increase j by 1), do:
         ans := maximum of ans and dp[j]
if fl is same as 0, then:
   return -1
Otherwise
   return ans + 1

Example

Let us see the following implementation to get better understanding −

#include <bits/stdc++.h>
using namespace std;
int solve(vector<vector<int>> matrix){
   int dp[210];
   int n = matrix.size();
   int fl = 1;
   int ans = 0;
   for (int i = 0; i < n && fl; i++){
      memset(dp, -1, sizeof(dp));
      dp[i] = 0;
      queue<int> q;
      q.push(i);
      while (!q.empty()){
         int x = q.front();
         q.pop();
         for (int j = 0; j < n; j++){
            if (matrix[x][j] == 1){
               if (dp[j] == -1){
                  dp[j] = dp[x] + 1;
                  q.push(j);
               }
               else if (abs(dp[j] - dp[x]) != 1)
                  fl = 0;
            }
         }
      }
      for (int j = 0; j < n; j++)
         ans = max(ans, dp[j]);
   }
   if (fl == 0){
      return -1;
   }else{
      return ans + 1;
   }
}
int main(){
   vector<vector<int>> matrix = { { 0, 1, 0, 1, 1, 0 }, { 1, 0, 1, 0, 0, 1 }, { 0, 1, 0, 1, 0, 0 }, { 1, 0, 1, 0, 0, 0 }, { 1, 0, 0, 0, 0, 0 }, { 0, 1, 0, 0, 0, 0 } };
   cout << solve(matrix) << endl;
}

Input

{ { 0, 1, 0, 1, 1, 0 }, { 1, 0, 1, 0, 0, 1 }, { 0, 1, 0, 1, 0, 0 }, { 1, 0, 1, 0, 0, 0 }, { 1, 0, 0, 0, 0, 0 }, { 0, 1, 0, 0, 0, 0 } }

Output

4

Arnab Chakraborty

e

Updated on: 02-Mar-2022

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