# C++ Program to Find Number of Articulation points in a Graph

Articulation Points (or Cut Vertices) in a Graph is a point iff removing it (and edges through it) disconnects the graph. An articulation point for a disconnected undirected graph, is a vertex removing which increases number of connected components.

## Algorithm

Begin
We use dfs here to find articulation point:
In DFS, a vertex w is articulation point if one of the following two conditions is satisfied.
1) w is root of DFS tree and it has at least two children.
2) w is not root of DFS tree and it has a child x such that no vertex in subtree rooted with
w has a back edge to one of the ancestors of w in the tree.
End

## Example

#include<iostream>
#include <list>
#define N -1
using namespace std;
class G {
int n;
//declaration of functions
void APT(int v, bool visited[], int dis[], int low[],
int par[], bool ap[]);
public:
G(int n); //constructor
void AP();
};
G::G(int n) {
this->n = n;
}
void G::addEd(int w, int x) {
}
void G::APT(int w, bool visited[], int dis[], int low[], int par[], bool ap[]) {
static int t=0;
int child = 0; //initialize child count of dfs tree is 0.
//mark current node as visited
visited[w] = true;
dis[w] = low[w] = ++t;
list<int>::iterator i;
int x = *i; //x is current adjacent
if (!visited[x]) {
child++;
par[x] = w;
APT(x, visited, dis, low, par, ap);
low[w] = min(low[w], low[x]);
// w is an articulation point in following cases :
// w is root of DFS tree and has two or more chilren.
if (par[w] == N && child> 1)
ap[w] = true;
// If w is not root and low value of one of its child is more than discovery value of w.
if (par[w] != N && low[x] >= dis[w])
ap[w] = true;
} else if (x != par[w]) //update low value
low[w] = min(low[w], dis[x]);
}
}
void G::AP() {
// Mark all the vertices as unvisited
bool *visited = new bool[n];
int *dis = new int[n];
int *low = new int[n];
int *par = new int[n];
bool *ap = new bool[n];
for (int i = 0; i < n; i++) {
par[i] = N;
visited[i] = false;
ap[i] = false;
}
// Call the APT() function to find articulation points in DFS tree rooted with vertex 'i'
for (int i = 0; i < n; i++)
if (visited[i] == false)
APT(i, visited, dis, low, par, ap);
//print the articulation points
for (int i = 0; i < n; i++)
if (ap[i] == true)
cout << i << " ";
}
int main() {
cout << "\nArticulation points in first graph \n";
G g1(5);
g1.AP();
return 0;
}

## Output

Articulation points in first graph
0 2

Updated on: 30-Jul-2019

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