#include <bits/stdc++.h>
#include "split.h"
using namespace std;
vector <vector <int> > adj;
vector <vector <int> > cadj;
vector <vector <int> > tadj;
vector <int> TVector;
int check[300003];
int SubtreeSize[300003];
int Parent[300003];
int n;
int N;
//Parent on each iteration of DFS
deque <int> clist;
vector <int> ans;
int csize;
int cback;
vector <int> Group1,Group2;
int GroupAID,GroupBID,GroupASize,GroupBSize;
int cur;
int Subtree1,Subtree2;
void ShittyDFS(int i){
cadj.clear();
adj=tadj;
for(int k=0;k<N;k++){
SubtreeSize[k]=0;
Parent[k]=0;
cadj.push_back(TVector);
}
Parent[i]=-1;
//wacky dfs (and calculate size of subtrees as well
//have to opt into a less retarded DFS to see whether things will work out or not.
//it does. good
cur=1;
clist.push_back(i);
check[i]=1;
while(!clist.empty()){
if(!SubtreeSize[clist.back()]){
SubtreeSize[clist.back()]=cur;
}
while(!adj[clist.back()].empty()&&check[adj[clist.back()].back()]){
adj[clist.back()].pop_back();
}
if(!adj[clist.back()].empty()){
check[adj[clist.back()].back()]=1;
Parent[adj[clist.back()].back()]=clist.back();
cback=clist.back();
cadj[clist.back()].push_back(adj[clist.back()].back());
cadj[adj[clist.back()].back()].push_back(clist.back());
clist.push_back(adj[cback].back());
cur++;
}else{
SubtreeSize[clist.back()]=cur-SubtreeSize[clist.back()]+1;
//while(!clist.empty()&&adj[clist.back()].empty()){
clist.pop_back();
//}
}
}
//retarded fucker deleted everything outta the graph after 1st instance of DFS. Has to DFS N times. Doesn't understand why the DFS refuses to work correctly
//It is, indeed, very highly recommended to NOT use IOI tasks to learn the basics :wheeze: :holyfuck:
}
vector <int> find_split(int n, int a, int b, int c, vector <int> p, vector <int> q){
//Only cares about the 2 smaller group: Assuming that we DID find a solution where the largest group is connected, just swap some of those points for the unused group.
//As this is the largest group, swapping some minor connected parts for the unused group is still valid, and always doable => only care about 2 smallest group only
//very intuitive tbh
N=n;
vector <pair <int,int> > order;
order.push_back({a,1});
order.push_back({b,2});
order.push_back({c,3});
sort(order.begin(),order.end());
GroupAID=order[0].second;
GroupASize=order[0].first;
GroupBID=order[1].second;
GroupBSize=order[1].first;
for(int i=0;i<N;i++){
adj.push_back(TVector);
}
//Build the graph. Who can't understand this?
for(int i=0;i<p.size();i++){
adj[p[i]].push_back(q[i]);
adj[q[i]].push_back(p[i]);
}
//Maybe random shuffle would work????
for(int i=0;i<n;i++){
// random_shuffle(adj[i].begin(),adj[i].end());
}
tadj=adj;
/*
Let's just call the smallest group gropu A, and second smallest group group B.
Let's kinda start working with the tree subtask first. When we split the tree into 2 part, we are essentially dedicating an entire subtree for group A and the rest of the
tree for group B, or vice versa. We will now do the following:\
- Try all vertices on the tree. For each verticies, check through all of its subtreees.
- If we can dedicate an entire subtree to group A and the rest to group B (or vice versa), this split will work instantly. We are basically cutting our original tree into 2 parts.
- If after checking all subtrees of all vertices and we still haven't found a solution, there is no solution.
(Why this kind of check works: We are trying everything, so no method of "splitting OG tree into 2 parts" is ignored...Let's be honest, I think that 64pts for this is trivial and
doesn't require much thought at all. The wacky funny parts comes from the final subtask, which I haven't solved yet.
Complexity: We take O(N) to DFS and calculate size of subtrees of all nodes. For each of the N vertex, we check through all of their adjacient vertices. Basically, we are checking
3N times (like, come on, in a tree of N nodes, there are 2N-2 pairs of adjacient points. There is literally nothing to explain here). Basically, for each DFS tree, we take roughly
O(N) to check whether there exists a valid split or not.
For the general graph (but small) case: Just create N DFS trees and bruteforce them, lol. Nothing to talk about just yet
*/
for(int i=0;i<N;i++){
//Tested locally, seems correct enough. First time implementing DFS. Moving on.
//On another hand, you know what? Chugging DFS into a function now. brb
//Done
ShittyDFS(i);
//Try the split with every single vertex on the tree.
for(int k=0;k<N;k++){
check[k]=0;
}
//Oh, one major note: The things below will have to be done on the DFS tree created on the previous step. Instead of BFS-ing on the original graph
//(which is the adj vector of edges), I will BFS on cadj (the graph that is the DFS tree created in this iteration)
for(int k=0;k<N;k++){
for(int l=0;l<cadj[k].size();l++){
Subtree1=SubtreeSize[cadj[k][l]];
if(cadj[k][l]==Parent[k]){
Subtree1=N-SubtreeSize[k];
}
Subtree2=N-Subtree1;
if(Subtree1>Subtree2){
swap(Subtree1,Subtree2);
//swap(GroupAID,GroupBID);
}
if(Subtree1>=GroupASize&&Subtree2>=GroupBSize){
//Found the answer so just BFS to mark the points
check[cadj[k][l]]=GroupAID;
check[k]=GroupBID;
clist.clear();
clist.push_back(cadj[k][l]);
csize=1;
while(csize<GroupASize){
for(int d=0;d<cadj[clist.front()].size();d++){
if(!check[cadj[clist.front()][d]]){
check[cadj[clist.front()][d]]=GroupAID;
clist.push_back(cadj[clist.front()][d]);
csize++;
if(csize==GroupASize){
break;
}
}
if(csize==GroupASize){
break;
}
}
clist.pop_front();
}
clist.clear();
clist.push_back(k);
csize=1;
while(csize<GroupBSize){
for(int d=0;d<cadj[clist.front()].size();d++){
if(!check[cadj[clist.front()][d]]){
check[cadj[clist.front()][d]]=GroupBID;
clist.push_back(cadj[clist.front()][d]);
csize++;
if(csize==GroupBSize){
break;
}
}
if(csize==GroupBSize){
break;
}
}
clist.pop_front();
}
ans.clear();
for(int d=0;d<N;d++){
if(check[d]==0){
check[d]=6-GroupAID-GroupBID;
}
ans.push_back(check[d]);
}
return ans;
}
//As it turns out, considering the reverse situation isn't needed. ok.
//Still can't figure out wtf is stopping this from passing subtask 4 for the life of me
}
}
if(p.size()==N-1){
//If it is a tree, the DFS tree is always the same, so 1 DFS is all it needs.
break;
}
}
ans.clear();
for(int i=0;i<N;i++){
ans.push_back(0);
}
return ans;
}
