# KPATHQRY - Editorial

Tester: Aryan
Editorialist: Taranpreet Singh

Easy-Medium

# PROBLEM

Given a tree with N nodes, we have to answer Q queries. Each query consists of K vertices, and we need to determine whether there exists a simple path in the tree, containing all these vertices.

# QUICK EXPLANATION

• Let us sort the vertices in the query in decreasing order of depth. Let u = V_1 and find v to be the first vertex in V such that v is not an ancestor of u.
• If no such v exists, then all vertices can lie on a single path.
• Otherwise, the vertices can lie on a simple path if and only if all vertices in query lie on a simple path from u to v, which can be checked easily with some preprocessing

# EXPLANATION

This problem has a variety of approaches to solve it. You can solve it by just using Euler tour and some basic observations, by building an auxiliary tree and checking if it’s a simple chain, or even using an overkill idea, maybe applying TALCA problem as a subproblem (just for fun)

I’d explain the idea I think is the simplest in terms of implementation.

The core idea is that we find a pair (u, v) such that we can claim that all vertices lie on a simple path if and only if they lie on the path from u to v.

Let us root the tree at any node. Now, consider a query, where set V denotes the vertices in the current query. Let’s pick the vertex u in V such that u is the deepest node present in set V.

Now, let’s pick another vertex v present in set V such that v is not an ancestor of u.

If there doesn’t exist such v, we can see that all nodes form a chain already, so the answer is YES.

Now, let’s assume we found such v.

Claim: All nodes in V can lie on the simple path if and only if all nodes lie on a path from u to v.
Proof: Assume tree is rooted at node R. Both u and v must lie on the path since they are in query. So, (u, v) is a candidate path. Let’s suppose L is the LCA of u and v. We have L \neq u and L \neq v. So, if we try to extend path (u, v), we would include nodes either in the subtree of node u or in the subtree of node v.

• No node in the subtree of u is included in the queried set, as u is the deepest node in the query set
• No node in the subtree of v is included in the queried set as if some node w in the subtree of node v was present in set V, then the depth of w is greater than the depth of v, so node w would have got chosen instead of node v.

Apart from node u and node v, all nodes on the path from node u to node v are connected to two vertices each, so we cannot include any other vertices, which are not already on the path from node u to node v.

Hence, the problem is reduced to checking that given path (u, v) and a set of vertices V, whether all vertices present in V lie on this path or not.

Unfortunately, the length of path (u, v) can be O(N), so we cannot just generate the whole path to answer each query. So we need to find a better way to check if some node w lie on path (u, v) or not.

Observation: If and only if node w lie on path from node u to node v, then dist(u, v) = dist(u, w) + dist(w, v).

This way, all we need to do is to optimize dist(a, b) queries which can be easily done by building euler tour of tree and computing LCA of nodes a and b, as we have dist(a, b) = depth_a + depth_b - 2*depth_L if L is LCA of a and b.

### Implementation

Hence, we need to build an Euler tour to both answer LCA queries, and to check if one node is the ancestor of the other or not. We also need to compute the depth of each vertex. Either use RMQ, or binary lifting to answer LCA queries.

### Bonus

Solve this problem using TALCA problem as a subproblem

Solving this problem using TALCA problem

Suppose the tree is rooted at node R.
For the query, find node v where node v is the deepest node present in queried nodes. Now, sort the vertices in V by their distance from v, and check if V_i is an ancestor of V_{i+1} when the tree is rooted at v.

# TIME COMPLEXITY

The time complexity is O(N*log(N) + \sum K*log(K)) or O((N + \sum K)*log(N)) or even (N*log(N) + \sum K) per test case depending upon implementation.

# SOLUTIONS

Setter's Solution
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
#define int long long int
#define ordered_set tree<int, nuint_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>
mt19937 rng(std::chrono::duration_cast<std::chrono::nanoseconds>(chrono::high_resolution_clock::now().time_since_epoch()).count());
#define mp make_pair
#define pb push_back
#define F first
#define S second
const int N=200005;
#define M 1000000007
#define BINF 1e9
#define init(arr,val) memset(arr,val,sizeof(arr))
#define MAXN 501
#define deb(xx) cout << #xx << " " << xx << "\n";
const int LG = 22;

int timer = 0, st[N], en[N], lvl[N], P[N][LG];

void dfs(int node, int parent) {
lvl[node] = 1 + lvl[parent];
P[node][0] = parent;

st[node] = timer++;
for (int i : adj[node]) {
if (i != parent) {
dfs(i, node);
}
}
en[node] = timer++;
}

void pre(int u, int p){
P[u][0] = p;
for(int i = 1; i < LG; i++)
P[u][i] = P[P[u][i - 1]][i - 1];

if(i != p)
pre(i, u);
}

int lca(int u, int v){
int i, lg;
if (lvl[u] < lvl[v]) swap(u, v);

for(lg = 0; (1<<lg) <= lvl[u]; lg++);
lg--;

for(i=lg; i>=0; i--){
if (lvl[u] - (1<<i) >= lvl[v])
u = P[u][i];
}

if (u == v)
return u;

for(i=lg; i>=0; i--){
if (P[u][i] != -1 and P[u][i] != P[v][i])
u = P[u][i], v = P[v][i];
}

return P[u][0];
}

}

void solve() {

int n;
cin >> n;

for(int i = 1; i <= n; i++){
st[i] = en[i] = lvl[i] = 0;
for(int j = 0; j < LG; j++){
P[i][j] = -1;
}
}
timer = 0;

for(int i = 0; i < n - 1; i++){
int x, y;
cin >> x >> y;
}

dfs(1, 0);
pre(1, 0);

int q;
cin >> q;

while(q--){

int k;
cin >> k;

assert(k >= 1 and k <= n);

vector<int> path(k);
for(int i = 0; i < k; i++){
cin >> path[i];
}

vector<pair<int, int>> v;
for(auto i : path){
v.push_back(make_pair(st[i], i));
}
sort(v.begin(), v.end());

vector<int>node;

for(int i = 0; i < k; i++){
bool got = false;
if(i + 1 < k and en[v[i + 1].S] < en[v[i].S]){
got = true;
}
if(!got){
node.push_back(v[i].S);
}
}

if(node.size() > 2){
cout << "NO\n";
continue ;
}
if(node.size() == 1){
cout << "YES\n";
continue ;
}

assert(node.size() == 2);

int lca_node = lca(node[0], node[1]);

int ok = 1;

for(auto i : path){
if(i != lca_node and lvl[lca_node] >= lvl[i]){
ok = 0;
break;
}
}

if(ok){
cout << "YES\n";
}else{
cout << "NO\n";
}

}

}

#undef int
int main() {
#define int long long int
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
#ifndef ONLINE_JUDGE
freopen("input.txt", "r", stdin);
freopen("optput.txt", "w", stdout);
#endif

int t;
cin >> t;
while(t--){
solve();
}

return 0;

}

Tester's Solution
/* in the name of Anton */

/*
Compete against Yourself.
Author - Aryan (@aryanc403)
Atcoder library - https://atcoder.github.io/ac-library/production/document_en/
*/

#ifdef ARYANC403
#else
#pragma GCC optimize ("Ofast")
#pragma GCC target ("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx")
//#pragma GCC optimize ("-ffloat-store")
#include<bits/stdc++.h>
#define dbg(args...) 42;
#endif

using namespace std;
#define fo(i,n)   for(i=0;i<(n);++i)
#define repA(i,j,n)   for(i=(j);i<=(n);++i)
#define repD(i,j,n)   for(i=(j);i>=(n);--i)
#define all(x) begin(x), end(x)
#define sz(x) ((lli)(x).size())
#define pb push_back
#define mp make_pair
#define X first
#define Y second
#define endl "\n"

typedef long long int lli;
typedef long double mytype;
typedef pair<lli,lli> ii;
typedef vector<ii> vii;
typedef vector<lli> vi;

const auto start_time = std::chrono::high_resolution_clock::now();
void aryanc403()
{
#ifdef ARYANC403
auto end_time = std::chrono::high_resolution_clock::now();
std::chrono::duration<double> diff = end_time-start_time;
cerr<<"Time Taken : "<<diff.count()<<"\n";
#endif
}

long long readInt(long long l, long long r, char endd) {
long long x=0;
cin>>x;return x;
int cnt=0;
int fi=-1;
bool is_neg=false;
while(true) {
char g=getchar();
if(g=='-') {
assert(fi==-1);
is_neg=true;
continue;
}
if('0'<=g&&g<='9') {
x*=10;
x+=g-'0';
if(cnt==0) {
fi=g-'0';
}
cnt++;
assert(fi!=0 || cnt==1);
assert(fi!=0 || is_neg==false);

assert(!(cnt>19 || ( cnt==19 && fi>1) ));
} else if(g==endd) {
if(is_neg) {
x=-x;
}
assert(l<=x&&x<=r);
return x;
} else {
assert(false);
}
}
}
string readString(int l, int r, char endd) {
string ret="";
int cnt=0;
while(true) {
char g=getchar();
assert(g!=-1);
if(g==endd) {
break;
}
cnt++;
ret+=g;
}
assert(l<=cnt&&cnt<=r);
return ret;
}
long long readIntSp(long long l, long long r) {
}
long long readIntLn(long long l, long long r) {
}
string readStringLn(int l, int r) {
}
string readStringSp(int l, int r) {
}

assert(getchar()==EOF);
}

vi a(n);
for(int i=0;i<n-1;++i)
return a;
}

const lli INF = 0xFFFFFFFFFFFFFFFL;

lli seed;
inline lli rnd(lli l=0,lli r=INF)
{return uniform_int_distribution<lli>(l,r)(rng);}

class CMP
{public:
bool operator()(ii a , ii b) //For min priority_queue .
{    return ! ( a.X < b.X || ( a.X==b.X && a.Y <= b.Y ));   }};

void add( map<lli,lli> &m, lli x,lli cnt=1)
{
auto jt=m.find(x);
if(jt==m.end())         m.insert({x,cnt});
else                    jt->Y+=cnt;
}

void del( map<lli,lli> &m, lli x,lli cnt=1)
{
auto jt=m.find(x);
if(jt->Y<=cnt)            m.erase(jt);
else                      jt->Y-=cnt;
}

bool cmp(const ii &a,const ii &b)
{
return a.X<b.X||(a.X==b.X&&a.Y<b.Y);
}

const lli mod = 1000000007L;
// const lli maxN = 1000000007L;

#include <algorithm>
#include <cassert>
#include <vector>

namespace atcoder {

struct dsu {
public:
dsu() : _n(0) {}
explicit dsu(int n) : _n(n), parent_or_size(n, -1) {}

int merge(int a, int b) {
assert(0 <= a && a < _n);
assert(0 <= b && b < _n);
if (x == y) return x;
if (-parent_or_size[x] < -parent_or_size[y]) std::swap(x, y);
parent_or_size[x] += parent_or_size[y];
parent_or_size[y] = x;
return x;
}

bool same(int a, int b) {
assert(0 <= a && a < _n);
assert(0 <= b && b < _n);
}

assert(0 <= a && a < _n);
if (parent_or_size[a] < 0) return a;
}

int size(int a) {
assert(0 <= a && a < _n);
}

std::vector<std::vector<int>> groups() {
for (int i = 0; i < _n; i++) {
}
std::vector<std::vector<int>> result(_n);
for (int i = 0; i < _n; i++) {
result[i].reserve(group_size[i]);
}
for (int i = 0; i < _n; i++) {
}
result.erase(
std::remove_if(result.begin(), result.end(),
[&](const std::vector<int>& v) { return v.empty(); }),
result.end());
return result;
}

private:
int _n;
std::vector<int> parent_or_size;
};

}  // namespace atcoder

#define rep(i, a, b) for(int i = a; i < (b); ++i)
#define trav(a, x) for(auto& a : x)
// #define all(x) begin(x), end(x)
// #define sz(x) (int)(x).size()
typedef long long ll;
typedef pair<int, int> pii;
// typedef vector<int> vi;

typedef vector<pii> vpi;
typedef vector<vpi> graph;

graph e(n);
atcoder::dsu d(n);
for(lli i=1;i<n;++i){
e[u].pb({v,1});
e[v].pb({u,1});
d.merge(u,v);
}
assert(d.size(0)==n);
return e;
}

template<class T>
struct RMQ {
vector<vector<T>> jmp;
RMQ(const vector<T>& V) {
int N = sz(V), on = 1, depth = 1;
while (on < N) on *= 2, depth++;
jmp.assign(depth, V);
rep(i,0,depth-1) rep(j,0,N)
jmp[i+1][j] = min(jmp[i][j],
jmp[i][min(N - 1, j + (1 << i))]);
}
T query(int a, int b) {
assert(a < b); // or return inf if a == b
int dep = 31 - __builtin_clz(b - a);
return min(jmp[dep][a], jmp[dep][b - (1 << dep)]);
}
};

struct LCA {
vi time;
vector<ll> dist;
RMQ<pii> rmq;

LCA(graph& C) : time(sz(C), -99), dist(sz(C)), rmq(dfs(C)) {}

vpi dfs(graph& C) {
vector<tuple<int, int, int, ll>> q(1);
vpi ret;
int T = 0, v, p, d; ll di;
while (!q.empty()) {
tie(v, p, d, di) = q.back();
q.pop_back();
if (d) ret.emplace_back(d, p);
time[v] = T++;
dist[v] = di;
trav(e, C[v]) if (e.first != p)
q.emplace_back(e.first, v, d+1, di + e.second);
}
return ret;
}

int query(int a, int b) {
if (a == b) return a;
a = time[a], b = time[b];
return rmq.query(min(a, b), max(a, b)).second;
}
ll distance(int a, int b) {
int lca = query(a, b);
return dist[a] + dist[b] - 2 * dist[lca];
}
};

vii buildEulerTour(const graph &e){
const lli n=sz(e);
lli tim=0;
vii tinout(n,mp(-1,-1));
function<void(lli,lli)> dfs2 = [&](lli u,lli p){
tinout[u].X=tim++;
for(auto x:e[u]){
if(p==x.X)
continue;
dfs2(x.X,u);
}
tinout[u].Y=tim++;
};

dfs2(0,-1);
return tinout;
}

int main(void) {
ios_base::sync_with_stdio(false);cin.tie(NULL);
// freopen("txt.in", "r", stdin);
// freopen("txt.out", "w", stdout);
// cout<<std::fixed<<std::setprecision(35);
lli sumN = 0;
while(T--)
{
sumN += n;
dbg(n);
dbg(sz(g));
dbg(q);
lli sumK = 0;
vector<bool> vis(n);
LCA lca(g);
auto tinout = buildEulerTour(g);
auto solveQuery=[&](vi &a){
if(sz(a)<=2)
return true;
lli lc=a[0];
for(auto x:a)
lc=lca.query(lc,x);
sort(all(a),[&](const lli x,const lli y){
return tinout[x].X<tinout[y].X;
});
reverse(all(a));
if(a.back()==lc)
a.pop_back();
const lli k=sz(a);
dbg(lc,a);
lli cnt=0;
for(lli i=0;i+1<k;++i){
const lli u=a[i],v=a[i+1];
const lli l=lca.query(u,v);
dbg(u,v,l);
if(l==lc){
if(cnt)
return false;
cnt++;
continue;
}
if(l!=v)
return false;
}
return true;
};

while(q--){
dbg(k);
sumK += k;
dbg(sz(a));
for(auto &x:a)
{
x--;
assert(!vis[x]);
vis[x]=true;
}
for(auto &x:a)
vis[x]=false;
cout<<(solveQuery(a)?"YES":"NO")<<endl;
dbg("query over");
}
// assert(sumK<=1e6);
}
// assert(sumN<=2e5);
aryanc403();
return 0;
}

Editorialist's Solution
import java.util.*;
import java.io.*;
class KPATHQRY{
//SOLUTION BEGIN
void pre() throws Exception{}
void solve(int TC) throws Exception{
int N = ni();
int[] from = new int[N-1], to = new int[N-1];
for(int i = 0; i< N-1; i++){
from[i] = ni()-1;
to[i] = ni()-1;
}
int[][] tree = tree(N, from, to);
LCA lca = new LCA(tree);
int[] dep = new int[N], st = new int[N], en = new int[N];
count = -1;
dfs(tree, dep, st, en, 0, -1);

for(int Q = ni(); Q> 0; Q--){
int K = ni();
Integer[] V = new Integer[K];
for(int i = 0; i< K; i++)V[i] = ni()-1;
Arrays.sort(V, (Integer i1, Integer i2) -> Integer.compare(dep[i1], dep[i2]));
int u = V[K-1], v = -1;
for(int i = K-2; i>= 0; i--){
if(st[V[i]] <= st[V[K-1]] && en[V[K-1]] <= en[V[i]])continue;
v = V[i];
break;
}
if(v == -1){
pn("YES");
continue;
}
//Only path from u to v can be good, so we check if all vertices lie on this path or not
boolean good = true;
int dist = lca.dist(u, v);
for(int x:V)good &= lca.dist(u, x)+lca.dist(x, v) == dist;
pn(good?"YES":"NO");
}
}
int count;
void dfs(int[][] tree, int[] dep, int[] st, int[] en, int u, int p){
st[u] = ++count;
for(int v:tree[u])if(v != p){
dep[v] = dep[u]+1;
dfs(tree, dep, st, en, v, u);
}
en[u] = count;
}
int[][] tree(int N, int[] from, int[] to){
int[] cnt = new int[N];
for(int x:from)cnt[x]++;
for(int x:to)cnt[x]++;
int[][] g = new int[N][];
for(int i = 0; i< N; i++)g[i] = new int[cnt[i]];
for(int i = 0; i< N-1; i++){
g[from[i]][--cnt[from[i]]] = to[i];
g[to[i]][--cnt[to[i]]] = from[i];
}
return g;
}
class LCA{
int n = 0, ti= -1;
int[] eu, fi, d;
RMQ rmq;
public LCA(int[][] g){
n = g.length;
eu = new int[2*n-1];fi = new int[n];d = new int[n];
Arrays.fill(fi, -1);Arrays.fill(eu, -1);
dfs(g, 0, -1);
rmq = new RMQ(eu, d);
}
public LCA(int[] eu, int[] fi, int[] d){
this.n = eu.length;
this.eu = eu;
this.fi = fi;
this.d = d;
rmq = new RMQ(eu, d);
}
void dfs(int[][] g, int u, int p){
eu[++ti] = u;fi[u] = ti;
for(int v:g[u])if(v!=p){
d[v] = d[u]+1;
dfs(g, v, u);eu[++ti] = u;
}
}
int lca(int u, int v){return rmq.query(Math.min(fi[u], fi[v]), Math.max(fi[u], fi[v]));}
int dist(int u, int v){return d[u]+d[v]-2*d[lca(u,v)];}
class RMQ{
int[] len, d;
int[][] rmq;
public RMQ(int[] ar, int[] weight){
len = new int[ar.length+1];
this.d = weight;
for(int i = 2; i<= ar.length; i++)len[i] = len[i>>1]+1;
rmq = new int[len[ar.length]+1][ar.length];
for(int i = 0; i< rmq.length; i++)
for(int j = 0; j< rmq[i].length; j++)
rmq[i][j] = -1;
for(int i = 0; i< ar.length; i++)rmq[0][i] = ar[i];
for(int b = 1; b<= len[ar.length]; b++)
for(int i = 0; i + (1<<b)-1< ar.length; i++)
if(weight[rmq[b-1][i]]<weight[rmq[b-1][i+(1<<(b-1))]])rmq[b][i] =rmq[b-1][i];
else rmq[b][i] = rmq[b-1][i+(1<<(b-1))];
}
int query(int l, int r){
if(l==r)return rmq[0][l];
int b = len[r-l];
if(d[rmq[b][l]]<d[rmq[b][r-(1<<b)]])return rmq[b][l];
return rmq[b][r-(1<<b)];
}
}
}
//SOLUTION END
void hold(boolean b)throws Exception{if(!b)throw new Exception("Hold right there, Sparky!");}
static boolean multipleTC = true;
void run() throws Exception{
out = new PrintWriter(System.out);
//Solution Credits: Taranpreet Singh
int T = (multipleTC)?ni():1;
pre();for(int t = 1; t<= T; t++)solve(t);
out.flush();
out.close();
}
public static void main(String[] args) throws Exception{
new KPATHQRY().run();
}
int bit(long n){return (n==0)?0:(1+bit(n&(n-1)));}
void p(Object o){out.print(o);}
void pn(Object o){out.println(o);}
void pni(Object o){out.println(o);out.flush();}
String n()throws Exception{return in.next();}
String nln()throws Exception{return in.nextLine();}
int ni()throws Exception{return Integer.parseInt(in.next());}
long nl()throws Exception{return Long.parseLong(in.next());}
double nd()throws Exception{return Double.parseDouble(in.next());}

StringTokenizer st;
}

}

String next() throws Exception{
while (st == null || !st.hasMoreElements()){
try{
}catch (IOException  e){
throw new Exception(e.toString());
}
}
return st.nextToken();
}

String nextLine() throws Exception{
String str = "";
try{
}catch (IOException e){
throw new Exception(e.toString());
}
return str;
}
}
}


Feel free to share your approach. Suggestions are welcomed as always.

6 Likes

My solution did not get accepted.

### Approach

I chose a node (say 1) and did depth first search from this node and save preprocess and post process time for each node starting with 1 for node 1.
Now I have time intervals for each node in which it was processed.

Now for every pair of interval, either they are disjoint or one is fully inside the other.
So, I will count number of pairs of disjoint intervals named cnt

Two intervals will be checked for disjoint-ness if they have same parent according to DFS applied before.
Now there are three cases

1. cnt > 1, those nodes cannot be on the simple path.
2. cnt = 0, those nodes are on the simple path.
3. cnt = 1, if those disjoint intervals are on first two levels then the nodes exist on simple path otherwise there is no simple path covering those nodes.
• Here, level is decides after extracting the intervals for query nodes and sorting them.

Help me find the problem in this approach.

My approach.

1. Pre-compute binary lifting and LCA
2. In every query, for given nodes, find the diameter (containing current query nodes only) and diameter end points. Let end points be x and y, length of diameter = d
3. For every node, We can easily find whether it lies on diameter. Let p = distance of node from diameter (think of it like perpendicular distance of point from line), We can see that p = (dist(x, cur_node) + dist(y, cur_node)) - diameter;
Hence if p = 0 then cur_node lies on diameter.
So if it is a simple path, every node in query must have p = 0;

Time complexity = O(T.(N.log(N) + Q.K.log(N)))
Check out my code: Solution: 48606187 | CodeChef

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First of all, your solution will give you TLE because you are checking the interval for every pair of elements in each query and the maximum number of elements in a query can be 10^5.
Secondly, Consider the following test case :
5 1
1 2
2 3
1 4
4 5
4 2 3 4 5

There are a total of 4 disjoint intervals so your solution will return false but these 4 vertices do lie on a simple path.

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“If you’re running out of problems, just copy a bunch of your past ones!”
https://www.codechef.com/COOK60/problems/KNODES

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Hello.
What is the formal condition to check if a node P lies in the simple path (U,V) ?

your test case is not tree, it has cycle 1 - 4 - 5

suppose intervals are [1,10], [2,9], [11,20], [12,19], there is one pair of disjoint interval [1,10] and [11,20]. I will not consider [2,9] with [11,20] or [12,19] since their parent are different in DFS.

checking for disjoint-ness can be done linearly in terms of number of nodes in query.

we just have to see if current interval (except first one) is outside the previous one. Every interval will be checked either with its parent or its sibling.

Solution: 48583452 | CodeChef
Any idea why its giving TLE in two test case

I tried the following solution but failed----
count = 0;
If a node contains an element from the array, it returns 1
if the both child of a node returns 1, count++
I thought on count > 1, it should not be a simple path because if more than 1 node contains the elements on the both side, it would create branches so Simple path could not be formed But it was a wrong approach.

i’ve seen this problem on geeksforgeeks, Its mentioned hard , even there. so to mark this problem as easy to medium, is truly de-motivating. Its not good for us new comers , so please make ammends to your rating system.

dist(U, P)+dist(P, V) = dist(U, V) holds if and only if P is on simple path from U to V

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I’ve also faced the same problem
[Solution: 48755135]

sir i apply your approch but i got TLE in my test cases
please tell me where i am wrong
here is my code Solution: 48755240 | CodeChef

while(node1!=0){
visited[node1]=1;
node1=parent[node1];
}
bec of this part u r getting TLE(I think). OR this one

while(node2!=0){
if(visited[node2]){
break;
}
visited[node2]=1;
node2=parent[node2];
}

We can also solve this problem using LCA, tree flattening and sorting.
First flatten the tree and note the in_time during dfs for each node.

Calculate the LCA of all query nodes, let’s name it L. Let’s call a child of L as special if it’s subtree contains at least one query node. If the count of special children is greater than two, answer is NO.
Next, sort all query nodes according to the increasing order of their in_time during dfs. Answer is YES if for every pair of adjacent nodes, their LCA is either one of the pair nodes or L.
https://www.codechef.com/viewsolution/48506723

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I took half a day to solve this by reading the editorial. Very well written. Can you give similar problems to solve?

Great editorial , i was confused how he was checking weather the node lie in the same chain or the end point of the different chain.

Hey, Found your approach quite fascinating. I did a similar approach. Sorted the query nodes based on the in_time and proceeded as follows:
If the 2nd node in the query is outside the subtree of the first node, set a variable limit as the out time of first node, else set it as the out time of the 2nd node. And as i traveresed the query from left to right, I only allowed to nodes in the current subtree, with an exception if the nodes escapes the current subtree as wells as the lmit variable set earlier, and this can happen only once. Now This approach is failing for the first 4 subtasks, and I am still not able to figure out the case where the algorithm could go wrong.

https://www.codechef.com/viewsolution/48613262

This is the code to my approach. A little help would be very appreciated.

Bro facing the same problem. Did the same approach, the first 4 subtasks are failing. Please let us know, if you come to find the problem.