I haven’t properly seen your code but it fails on
1
8 3
1 1 1 1 1 1 1 1
1 2
2 3
3 4
4 7
2 5
5 6
6 8
7
3
1
Your output - 4 1 0 1 0 1 0 1
Correct output - 5 1 0 1 0 1 0 0
I was coding segment tree along with euler tour. Only Update function was left when I realised one doesn’t need it. 
“Things to watch out for” Did got one TLE due to this.
2 Likes
What is wrong with this → CodeChef: Practical coding for everyone
Please help someone .I used BFS and created a distance array for each Node.
I am getting runtime array but it is working fine for the sample test case.
Thanks brother , I was missing a Q.pop() operation when I was using continue statement But after correcting it also I am getting correct answer on your test case but WA during submission. Can you now tell me some test cases where my solution is failing . Thanks in advance.
https://www.codechef.com/viewsolution/37007493
Actually your approach is not correct. We have to transfer the values of nodes at even distance to the source node “present in that subtree”.
But according to your approach you are transfering values of all nodes whose depth is even from that node, so there may be some nodes which can not be present in that subtree.
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" You are given a tree rooted at node 1 with N vertices."
This is the first line of the question .
3 Likes
Can somebody help me out finding what is wrong with this solution. Thanks in advance!
https://www.codechef.com/viewsolution/37003547
Can anybody explain me this…It is written in the editorial that “if visited[vertex][1] == 1, all the nodes at odd distance from the current vertex are zero. So there is no need to go further.” but, aren’t we interested in the nodes at even distance from the vertex so why no need to go further?
My Sol with just one simple DFS to do the work :)
`//
// author : Abhishek
//
#include <bits/stdc++.h>
#include
#define ll long long
#define db double
#define deb(x) cout<<#x<<“: “<<x<<”\n”
#define all(x) x.begin(),x.end()
#define fo(i, n) for(i=0 ; i<n ; i++)
#define Fo(i, k, n) for(i=k ; i<n ; i++)
#define setBits(x) __builtin_popcountll(x)
#define parity(x) __builtin_parity(x)
using namespace std;
using namespace chrono;
const ll MOD = 1000000007;
const ll N = 300005; // array size
const ll inf = 10000000000000000;
const db PI = 3.14159265358979323846264338327950288419716939937510;
void print(vector v) {
ll i;
fo(i, v.size())cout << v[i] << " ";
cout << “\n”;
}
void print(vector<pair<ll, ll> > v) {
ll i;
fo(i, v.size()) {
cout << v[i].first << " " << v[i].second << “\n”;
}
}
vector read(ll n) {
ll i;
vector arr(n, 0);
fo(i, n)cin >> arr[i];
return arr;
}
vector read(ll n, ll k) {
ll i;
vector arr(n + k, 0);
Fo(i, k, n + k)cin >> arr[i];
return arr;
}
bool isEven(ll n) {
return n % 2 == 0;
}
bool isOdd(ll n) {
return n % 2 != 0;
}
vector adj[N];
vector vis(N, 0);
vector taken(N, 0);
vector ans(N, 0);
vector depth(N, 0);
pair<ll, ll> dfs(ll cur, bool takenEv, bool takenOd) {
vis[cur] = true;
pair<ll, ll> forThis = {0, 0};
if (isEven(depth[cur])) {
forThis.first += ans[cur];
} else {
forThis.second += ans[cur];
}
for (auto next: adj[cur]) {
if (!vis[next]) {
depth[next] = 1 + depth[cur];
pair<ll, ll> ct;
if (taken[cur]) {
if (isEven(depth[cur])) {
ct = dfs(next, true, takenOd);
} else {
ct = dfs(next, takenEv, true);
}
} else {
ct = dfs(next, takenEv, takenOd);
}
forThis.first += ct.first;
forThis.second += ct.second;
}
}
if (takenEv && isEven(depth[cur])) {
ans[cur] = 0;
} else if (takenOd && isOdd(depth[cur])) {
ans[cur] = 0;
} else if (taken[cur]) {
ans[cur] = (isEven(depth[cur]) ? forThis.first : forThis.second);
}
return forThis;
}
void solve() {
ll t = 1, cs = 1;
cin >> (t);
while (t–) {
ll i, j, k, n, m, p, q, x, y, a, b, l, r;
cin >> n >> q;
Fo(i, 1, n + 1) {
adj[i].clear();
taken[i] = 0;
ans[i] = 0;
vis[i] = false;
}
Fo(i, 1, n + 1) {
cin >> ans[i];
}
fo(i, n - 1) {
cin >> x >> y;
adj[x].push_back(y);
adj[y].push_back(x);
}
while (q--) {
cin >> x;
taken[x] = 1;
}
dfs(1, 0, 0);
// Fo(i, 1, n + 1) {
// cout << depth[i] << " ";
// }
// cout << “\n”;
Fo(i, 1, n + 1) {
cout << ans[i] << " ";
}
cout << "\n";
}
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
#ifndef ONLINE_JUDGE
freopen(“input.txt”, “r”, stdin);
freopen(“output.txt”, “w”, stdout);
freopen(“error.txt”, “w”, stderr);
#endif
auto start = high_resolution_clock::now();
solve();
ll i, j, k, n, m, p, q, x, y, a, b, l, r;
auto stop = high_resolution_clock::now();
auto duration = duration_cast<microseconds>(stop - start);
#ifndef ONLINE_JUDGE
cerr << "Time: " << duration.count() / 1000.0 << endl;
#endif
return 0;
}
`
bro! anyone there who can help me . i have written code using euler tour but still showing tle. and i have checked it for many test cases and even given in the comments too.
here is my ideone link
Hello community
Mine approach is also similar, however I am still getting Wrong Answer. Can anyone please help me figure out my mistake.
My solution : CodeChef: Practical coding for everyone
Thank You 
I am getting WA for the below solution … i dont know whats wrong. please help!
https://www.codechef.com/viewsolution/37019870
Can someone explain how a star-shaped tree would result in a TLE?
Can anyone please tell me where my code fails… I cannot find it. I’m getting TLE on submission
https://www.codechef.com/viewsolution/37027990
https://www.codechef.com/viewsolution/37029733
My solution is almost same but I keep getting the please help
for each query,i made a dfs and added even distnce of subtree values to vertex.
As it is mentioned sum of all N and Q over all test cases doen’t exceed 200000 ,its no where near to get a TLE,but i did get TLE 
,can anyone look into my code and say what’s wrong?
https://www.codechef.com/viewsolution/37033454