My approach is while creating edge if anyone having black colour tranfer it to parent and while changing colour means query 2 just change colour of parent ,,, Problem
class DSU{
public:
vector<int>par;
vector<pair<int,int>>rnk;
// 0->white,1->black
DSU(int n){
par.resize(n+1);
rnk.assign(n+1,{1,0});
for(int i=1;i<=n;i++)par[i]=i;
}
int find_pita(int src){
if(src==par[src]) return src;
return par[src]=find_pita(par[src]);
}
void take_union(int a,int b){
a=find_pita(a),b=find_pita(b);
if(a==b)return ;
if(rnk[a].first>rnk[b].first){
rnk[a].first+=rnk[b].first;
par[b]=a;
if(rnk[b].second==1)rnk[a].second=1;
}else{
rnk[b].first+=rnk[a].first;
par[a]=b;
if(rnk[a].second==1)rnk[b].second=1;
}
}
void flip_col(int src){
int u=find_pita(src);
rnk[u].second=!rnk[u].second;
}
};
void coderaryan() {
// executing code from here
int t=1;
while (t--) {
int n,q;
cin >> n>>q;
DSU vab(n);
while(q--){
1 int x;cin>>x;
if(x==1){
int u,v;cin>>u>>v;
vab.take_union(u,v);
}else if(x==2){
int u;cin>>u;
vab.flip_col(u);
}else{
int u;cin>>u;
u=vab.find_pita(u);
vab.rnk[u].second==1?cout<<"Yes"<<endl:cout<<"No"<<endl;
}
}
}
}
I think the issue arises when a connected component contains more than one black vertex. If you flip one of those black vertices, the code prints “No,” even though at least one black vertex remains. For example:
At this point both vertices are black. Flipping either one reduces the count to one, yet the code outputs “No” instead of “Yes.”
To fix this, keep:
1- A color[u] array for each vertex’s current color.
2- A blackCount[find(u)] per component to track the exact count of blacks.
Now querying “is there any black?” is simply
(I reworked this comment with AI to make it clearer, since my English isn’t very good.)