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Check if a given graph is Bipartite using DFS in C++ program
Suppose we have a connected graph; we have to check whether the graph is bipartite or not. If the graph coloring is possible applying two colors such that nodes in a set are colored with the same color.
So, if the input is like
then the output will be True
To solve this, we will follow these steps −
- Define a function insert_edge(), this will take an edge array adj, u, v,
- insert v at the end of adj[u]
- insert u at the end of adj[v]
- From the main method do the following,
- for each u in adj[v],do
- if visited[u] is same as false, then −
- visited[u] := true
- color[u] := invert of color[v]
- if not is_bipartite_graph(adj, u, visited, color), then −
- return false
- otherwise when color[u] is same as color[v], then −
- return false
- if visited[u] is same as false, then −
- return true
Example (C++)
Let us see the following implementation to get better understanding −
#include <bits/stdc++.h>
using namespace std;
void insert_edge(vector<int> adj[], int u, int v){
adj[u].push_back(v);
adj[v].push_back(u);
}
bool is_bipartite_graph(vector<int> adj[], int v, vector<bool>& visited, vector<int>& color){
for (int u : adj[v]) {
if (visited[u] == false) {
visited[u] = true;
color[u] = !color[v];
if (!is_bipartite_graph(adj, u, visited, color))
return false;
}
else if (color[u] == color[v])
return false;
}
return true;
}
int main() {
int N = 6;
vector<int> adj_list[N + 1];
vector<bool> visited(N + 1);
vector<int> color(N + 1);
insert_edge(adj_list, 1, 2);
insert_edge(adj_list, 2, 3);
insert_edge(adj_list, 3, 4);
insert_edge(adj_list, 4, 5);
insert_edge(adj_list, 5, 6);
insert_edge(adj_list, 6, 1);
visited[1] = true;
color[1] = 0;
cout << (is_bipartite_graph(adj_list, 1, visited, color));
}Input
insert_edge(adj_list, 1, 2); insert_edge(adj_list, 2, 3); insert_edge(adj_list, 3, 4); insert_edge(adj_list, 4, 5); insert_edge(adj_list, 5, 6); insert_edge(adj_list, 6, 1);
Output
1
Updated on: 27-Aug-2020
506 Views
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