# Find two disjoint good sets of vertices in a given graph

Given an undirected unweighted graph with N vertices and M edges. The task is to find two disjoint good sets of vertices. A set X is called good if for every edge UV in the graph at least one of the endpoint belongs to X(i.e, U or V or both U and V belongs to X).
If it is not possible to make such sets then print -1.

Examples:

Input : Output : {1 3 4 5} ,{2 6}
One disjoint good set contains vertices {1, 3, 4, 5} and other contains {2, 6}.

Input : Output : -1

Approach:
One of the observation is that there is no edge UV that U and V are in the same set.The two good sets form a bipartition of the graph, so the graph has to be bipartite. And being bipartite is also sufficient. Read about bipartition here.

Below is the implementation of the above approach :

 `// C++ program to find two disjoint ` `// good sets of vertices in a given graph ` `#include ` `using` `namespace` `std; ` `#define N 100005 ` ` `  `// For the graph ` `vector<``int``> gr[N], dis; ` `bool` `vis[N]; ` `int` `colour[N]; ` `bool` `bip; ` ` `  `// Function to add edge ` `void` `Add_edge(``int` `x, ``int` `y) ` `{ ` `    ``gr[x].push_back(y); ` `    ``gr[y].push_back(x); ` `} ` ` `  `// Bipartie function ` `void` `dfs(``int` `x, ``int` `col) ` `{ ` `    ``vis[x] = ``true``; ` `    ``colour[x] = col; ` ` `  `    ``// Check for child vertices ` `    ``for` `(``auto` `i : gr[x]) { ` ` `  `        ``// If it is not visited ` `        ``if` `(!vis[i]) ` `            ``dfs(i, col ^ 1); ` ` `  `        ``// If it is already visited ` `        ``else` `if` `(colour[i] == col) ` `            ``bip = ``false``; ` `    ``} ` `} ` ` `  `// Function to find two disjoint ` `// good sets of vertices in a given graph ` `void` `goodsets(``int` `n) ` `{ ` `    ``// Initially assume that graph is bipartie ` `    ``bip = ``true``; ` ` `  `    ``// For every unvisited vertex call dfs ` `    ``for` `(``int` `i = 1; i <= n; i++) ` `        ``if` `(!vis[i]) ` `            ``dfs(i, 0); ` ` `  `    ``// If graph is not bipartie ` `    ``if` `(!bip) ` `        ``cout << -1; ` `    ``else` `{ ` ` `  `        ``// Differentiate two sets ` `        ``for` `(``int` `i = 1; i <= n; i++) ` `            ``dis[colour[i]].push_back(i); ` ` `  `        ``// Print vertices belongs to both sets ` ` `  `        ``for` `(``int` `i = 0; i < 2; i++) { ` ` `  `            ``for` `(``int` `j = 0; j < dis[i].size(); j++) ` `                ``cout << dis[i][j] << ``" "``; ` `            ``cout << endl; ` `        ``} ` `    ``} ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `n = 6, m = 4; ` ` `  `    ``// Add edges ` `    ``Add_edge(1, 2); ` `    ``Add_edge(2, 3); ` `    ``Add_edge(2, 4); ` `    ``Add_edge(5, 6); ` ` `  `    ``// Function call ` `    ``goodsets(n); ` `} `

Output:

```1 3 4 5
2 6
```

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