In a company's OA, I came across the question of counting all cycles in a graph and used the coloring method. Can anybody tell me how to improvise this solution so I can count a cycle that includes a node that's part of more than two cycles?

void dfs(int s,int p,vector<int> &color,vector<int> &parent){
    if(color[s]==2)return;
    if(color[s]==1){
        int cur=p;
        vector<int> vec;
        ans++;                // count of cycles
        vec.push_back(cur);
        while(cur!=s){
            cur=parent[cur];
            vec.push_back(cur);
        }
        cycles.push_back(vec); // to store answer
        return ;
    }
    parent[s]=p;
    color[s]=1;
    for(auto &val:graph[s]){
        if(val!=p){
            dfs(val,s,color,parent);
        }
    }
    color[s]=2;
    return;
}

Output for addEdge(1,4);
addEdge(1,2); addEdge(1,3); addEdge(3,4); addEdge(2,3); is {3, 4, 1} and {2, 3, 4, 1}.

but for this one addEdge(1,2); addEdge(1,4); addEdge(1,3); addEdge(3,4); addEdge(2,3); is {3, 2, 1} and {4, 3, 2, 1}.

I just changed the order. Nothing else graph's structure is still the same.