# Minimize the number of weakly connected nodes

Given an **undirected graph**, task is to find the minimum number of weakly connected nodes after converting this graph into directed one.

**Weakly Connected Nodes :** Nodes which are having 0 indegree(number of incoming edges).

**Prerequisite :** BFS traversal

Examples :

Input : 4 4 0 1 1 2 2 3 3 0 Output : 0 disconnected components Input : 6 5 1 2 2 3 4 5 4 6 5 6 Output : 1 disconnected components

**Explanation :**

**Approach :** We find a node which helps in traversing maximum nodes in a single walk. To cover all possible paths, DFS graph traversal technique is used for this.

Do the above steps to traverse the graph. Now, iterate through graph again and check which nodes are having 0 indegree.

`// CPP code to minimize the number ` `// of weakly connected nodes ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Set of nodes which are traversed ` `// in each launch of the DFS ` `set<` `int` `> node; ` `vector<` `int` `> Graph[10001]; ` ` ` `// Function traversing the graph using DFS ` `// approach and updating the set of nodes ` `void` `dfs(` `bool` `visit[], ` `int` `src) ` `{ ` ` ` `visit[src] = ` `true` `; ` ` ` `node.insert(src); ` ` ` `int` `len = Graph[src].size(); ` ` ` `for` `(` `int` `i = 0; i < len; i++) ` ` ` `if` `(!visit[Graph[src][i]]) ` ` ` `dfs(visit, Graph[src][i]); ` `} ` ` ` `// building a undirected graph ` `void` `buildGraph(` `int` `x[], ` `int` `y[], ` `int` `len){ ` ` ` ` ` `for` `(` `int` `i = 0; i < len; i++) ` ` ` `{ ` ` ` `int` `p = x[i]; ` ` ` `int` `q = y[i]; ` ` ` `Graph[p].push_back(q); ` ` ` `Graph[q].push_back(p); ` ` ` `} ` `} ` ` ` `// computes the minimum number of disconnected ` `// components when a bi-directed graph is ` `// converted to a undirected graph ` `int` `compute(` `int` `n) ` `{ ` ` ` `// Declaring and initializing ` ` ` `// a visited array ` ` ` `bool` `visit[n + 5]; ` ` ` `memset` `(visit, ` `false` `, ` `sizeof` `(visit)); ` ` ` `int` `number_of_nodes = 0; ` ` ` ` ` `// We check if each node is ` ` ` `// visited once or not ` ` ` `for` `(` `int` `i = 0; i < n; i++) ` ` ` `{ ` ` ` `// We only launch DFS from a ` ` ` `// node iff it is unvisited. ` ` ` `if` `(!visit[i]) { ` ` ` ` ` `// Clearing the set of nodes ` ` ` `// on every relaunch of DFS ` ` ` `node.clear(); ` ` ` ` ` `// relaunching DFS from an ` ` ` `// unvisited node. ` ` ` `dfs(visit, i); ` ` ` ` ` `// iterating over the node set to count the ` ` ` `// number of nodes visited after making the ` ` ` `// graph directed and storing it in the ` ` ` `// variable count. If count / 2 == number ` ` ` `// of nodes - 1, then increment count by 1. ` ` ` `int` `count = 0; ` ` ` `for` `(` `auto` `it = node.begin(); it != node.end(); ++it) ` ` ` `count += Graph[(*it)].size(); ` ` ` ` ` `count /= 2; ` ` ` `if` `(count == node.size() - 1) ` ` ` `number_of_nodes++; ` ` ` `} ` ` ` `} ` ` ` `return` `number_of_nodes; ` `} ` ` ` `//Driver function ` `int` `main() ` `{ ` ` ` `int` `n = 6,m = 4; ` ` ` `int` `x[m + 5] = {1, 1, 4, 4}; ` ` ` `int` `y[m+5] = {2, 3, 5, 6}; ` ` ` ` ` `/*For given x and y above, graph is as below : ` ` ` `1-----2 4------5 ` ` ` `| | ` ` ` `| | ` ` ` `| | ` ` ` `3 6 ` ` ` ` ` `// Note : This code will work for ` ` ` `// connected graph also as : ` ` ` `1-----2 ` ` ` `| | \ ` ` ` `| | \ ` ` ` `| | \ ` ` ` `3-----4----5 ` ` ` `*/` ` ` ` ` `// Building graph in the form of a adjacency list ` ` ` `buildGraph(x, y, n); ` ` ` `cout << compute(n) << ` `" weakly connected nodes"` `; ` ` ` ` ` `return` `0; ` `} ` |

*chevron_right*

*filter_none*

**Output:**

2 weakly connected nodes

## Recommended Posts:

- Kth largest node among all directly connected nodes to the given node in an undirected graph
- Number of connected components in a 2-D matrix of strings
- Number of ways to select a node from each connected component
- Program to count Number of connected components in an undirected graph
- Maximum number of edges among all connected components of an undirected graph
- Print levels with odd number of nodes and even number of nodes
- Count the nodes whose sum with X is a Fibonacci number
- Count the number of non-reachable nodes
- Number of sink nodes in a graph
- Calculate number of nodes in all subtrees | Using DFS
- Number of special nodes in an n-ary tree
- Number of Unicolored Paths between two nodes
- Count the number of nodes at a given level in a tree using DFS
- Count the number of nodes at given level in a tree using BFS.
- Level with maximum number of nodes using DFS in a N-ary tree

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.