Auto comment: topic has been updated by tkien1707 (previous revision, new revision, compare).

Can you just share link of the problem? I will try to solve it

The only part of this problem that differs from the main DSU structure is the "NOT GIVEN" response.

For "YES" and "NO" just use a very basic DSU (or with some simple modifications, if you like), and for "NOT GIVEN" use a map/set/etc. where you will store information if any number already appeared in type 1 query, and if it's not then it's the only case when "NOT GIVEN" is a valid response (and you need to check for it BEFORE the DSU query). That's all.

Update: I checked the original problem statement, the solution I described above won't work for this (consider 1 1 2 and 1 3 4 and 2 1 4 — the response should be "NOT GIVEN" even though these numbers have already appeared). I believe that this can be solved using bipartite graph modification for the type 1 query, where if the both ingridients are unknown you create new bipartite group with them, and if one of them is known then you merge it. If in type 2 query two ingridients are from different groups then it's "NOT GIVEN", if they are from different sets the answer is "NO", otherwise "YES"

For type 1 query add an edge between 2 nodes and color them opposite color
For type 2 if nodes belong to different component we can't determine otherwise if they have same color it is safe and fatal if different color.
While merging 2 components, merge small to large

map<int,set<int>>mp;
vector<int>par(1e5+10),val(1e5+10);
int find_par(int v){
    if(v==par[v]){
        return v;
    }
    mp[par[v]].erase(v);
    par[v]=find_par(par[v]);
    mp[par[v]].insert(v);
    return par[v];
}
void merge(int u,int v){
    int pu=find_par(u);
    int pv=find_par(v);
    if(pu==pv){
        return;
    }
    if(mp[pu].size()>mp[pv].size()){
        swap(pu,pv);
        swap(u,v);
    }
    if(val[u]==val[v]){
        for(int x:mp[pu]){
            val[x]=1-val[x];
            mp[pv].insert(x);
        }
        val[pu]=1-val[pu];
        mp[pv].insert(pu);
    }
    mp[pu].clear();
    par[pu]=pv;
}
void solve(int ttc){
    int n,q;
    cin>>n>>q;
    for(int i=0;i<n;i++){
        par[i]=i;
        val[i]=1;
    }
    while(q--){
        int t,u,v;
        cin>>t>>u>>v;
        u--;v--;
        if(t==2){
            int pu=find_par(u);
            int pv=find_par(v);
            if(pu!=pv){
                cout<<"DUNNO\n";
            }else{
                if(val[u]==val[v]){
                    cout<<"SAFE\n";
                }else{
                    cout<<"FATAL\n";
                }
            }
        }else{
            merge(u,v);
        }
    }
}