 # Having doubts and Getting wrong answer for ELWINTER problem from MARCH LUNCHTIME

i have copy pasted this code from editorial , but it is giving me wrong answer for the given test cases and i am also having some doubts regarding the approach which is discussed in the editorial . Can Somebody Help?

Doubt
why do we need two queues i.e. `qq` and `q`? what is their purpose?

Code:-

``````#include "bits/stdc++.h"
#define ll long long int
#define MOD 1000000007
using namespace std;

void frozeAdj(vector<vector<ll>> &graph, queue<ll> &q, vector<ll> &arr)
{
ll rep = q.size();
while (rep--)
{
auto i = q.front();
q.pop();
if (arr[i])
continue;
arr[i] = 1;

for (ll j : graph[i])
q.push(j);
}
}

int main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
ll t;
cin >> t;
while (t--)
{
ll n, m, query;
vector<ll> arr(n + 1, 0); // array for tracing if node is frozen
queue<ll> q;              // to implement frozeAdjacent function
vector<vector<ll>> graph(n + 1);
cin >> n >> m >> query;

for (ll i = 0; i < m; i++)
{
ll a, b;
cin >> a >> b;
graph[a].push_back(b);
graph[b].push_back(a);
}

ll mxtimes = n - 1;
queue<ll> qq;
while (query--)
{
ll a, b;
cin >> a >> b;
if (a == 1)
{
if (!arr[b])
{
qq.push(b);
arr[b] = 1;
}
}

else if (a == 2)
{
while (!qq.empty())
{
ll i = qq.front();
qq.pop();
for (ll j : graph[i])
q.push(j);
}

if (q.empty() || mxtimes == 0)
continue;

while (b-- && mxtimes > 0)
{
mxtimes--;
}
}

else
{
if (arr[b])
cout << "YES" << endl;
else
cout << "NO" << endl;
}
}
}
return 0;
}
``````

The two queues are essential for the proper state-transition from time T to time T+1.

Check the following accepted code .

Accepted

Can you tell why is the Dijkstra based solution failing ? All that was done is , added. extra edges from source node(node which was frozen before all others) to other nodes which are getting froze at later timings , and edge weight = (time of source node - time of other node) , solution is failing : Solution: 60943047 | CodeChef

After that , I tried the simple bfs solution , which passed (passed) all the cases. Any idea why dijkstra solution fails ? @codingknight

I am not able to understand

Check the following update to your Dijkstra based solution.

Accepted