# Applications of DFS

One of the applications of DFS is to find the number of connected components in a graph.In graph theory, a connected component of an undirected graph is a sub-graph in which any two vertices are connected to each other by paths, and which is connected to no additional vertices in the super-graph.

Algorithm-

A depth first search that begins at some particular vertex v will find the entire connected component containing v (and no more) before returning.To find all the connected components of a graph, loop through its vertices, starting a new depth first search whenever the loop reaches a vertex that has not already been included in a previously found connected component.

Another application is to find if the given graph is a tree or not.A tree is an undirected graph which is connected and does not have cycles.

Properties of a tree-

1.The tree should have only one connected component ie. it is not a disconnected graph.
2.If the graph has N vertices and N-1 edges,then it is a tree.This condition ensures that no cycles are formed.

Algorithm-

For a graph to be a tree,both the above conditions must be satisfied.Number of components is found using the above algorithm and it should be 1.Then,the second relation must be verified.

Code-

```
#include<stdio.h>
#include<vector>
using namespace std;

vector<int>arr;
int n,a,b,j=0,m;
int color;

//counter counts the number of components

int counter;

void dfs(int node)
{
color[node]=2;//marking the visited nodes

for(int i=0;i<arr[node].size();++i)
{

if(color[arr[node][i]]==0)
{
dfs(arr[node][i]);
}
}
}

int main()
{
int t,p;
int i;
counter=0;
p=1;
j=0;

// n is the number of vertices
// m is the number of edges

scanf("%d %d",&n,&m);

//clearing the arrays and vectors

for(i=0;i<n;i++)
arr[i].clear();

for(i=0;i<n;++i){
color[i]=0;
}

while(j<m){

//a and b denote the starting and ending points of the edge

scanf("%d %d",&a,&b);

arr[a-1].push_back(b-1);
arr[b-1].push_back(a-1);
j++;

}
//looping through the vertices

for(i=0;i<n;++i)
{
if(color[i]==0)
{
dfs(i);
counter++;
}
}

printf("Number of components is %d\n",counter);

if((m==n-1)&&(counter==1))printf("IT IS A TREE\n");

else printf("NOT A TREE\n");

return 0;
}
```

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