Count the paths

PROBLEM :

Given a directed graph, a source vertex ‘s’ and a destination vertex ‘d’, print the count of all paths from given ‘s’ to ‘d’.

Input:
First line consists of T test cases. First line of every test case consists of V and E, denoting the vertices and edges. Second line of every test case consists of 2*E spaced integers denoting the edge between vertices. Third line of every test case consists of S and D.

Output:
Single line output, print the count of all paths from 's' to 'd'.

Constraints:
1<=T<=100
1<=V,E<=10

Example:
Input:
1
4 6
0 1 0 2 0 3 2 0 2 1 1 3
2 3

Output:
3

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SIMPLE c++ IMPLEMENTATION :
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#include<iostream>
using namespace std;
#include <list>

class graph
{
    int v ;
    list<int> *adj ;
   
    public :
        graph(int v) ;
        void addEdges(int s,int d) ;
        int CountPaths(int s,int d) ;
        void UtilCountPaths(int s,int d,bool visited[],int *count) ;
} ;

graph::graph(int v)
{
    this->v=v ;
    adj=new list<int>[v] ;
}

void graph:: addEdges(int s,int d)
{
    adj[s].push_back(d) ;
}

int graph:: CountPaths(int s,int d)
{
    bool visited[v] ;
   
    for(int i=0;i<v;i++)
        visited[i]=false ;
       
    int count=0 ;
   
    UtilCountPaths(s,d,visited,&count) ;
    return count ;
}

void graph:: UtilCountPaths(int s,int d,bool visited[],int *count)
{
    visited[s]=true ;
   
    if(s==d)
        (*count)++ ;
       
    else
    {
        list<int>::iterator i ;
        for(i=adj[s].begin();i!=adj[s].end();i++)
        {
            if(!visited[*i])
                UtilCountPaths(*i,d,visited,&(*count)) ;
        }
    }
    visited[s]=false ;
}

int main()
{
int t ;
cin>>t ;
while(t--)
{
   int v,e ;
   cin>>v>>e ;
 
   graph g(v) ;
 
   int s,d ;
   for(int i=0;i<e;i++)
   {
       cin>>s>>d ;
       g.addEdges(s,d) ;
   }
 
   cin>>s>>d ;
   int path=g.CountPaths(s,d) ;
   cout<<path<<endl ;
}
return 0;
}

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