It really feels like A<B<C<D<F<<G<<<E

Bring back 'Aoki' and 'Takahashi'.

I got cooked in this contest.

How to do F?

Can someone please explain E problem?

In my opinion, the order of the problems shouble be ABCDFGE

first time i solved F in an atcoder contest, very nice contest!

Problem E is so impressive !

At first, I thought that it is based on some algorithm associated with graph, like Dijkstra or something else, but kept getting WA or TLE.

Then, I convinced myself that "solving it based on graph algorithm" is just my mindset, which in fact I didn't have any proof.

Finally, I came up with another solution, which almost has nothing to do with graph. Note that a train TR1 can only be affected by another train TR2, if TR2's T <= TR1's S, where S and T are the time mentioned in the problem. So, we can deal with the "time" in an increasing order. For each time, at first, we deal with the trains that arrive at this time, and then deal with the trains that leave at this time. For any train that arrives at some city at this time, we update the maximum "arriving time" of this city, and then for any train that leaves some city at this time, we update its minimum "delay time" according to the city's maximum "arriving time". Note that as we handle everything in an increasing order of the time, the solution is always correct.

can anyone tell why this code gives runtime error for two cases:

https://atcoder.jp/contests/abc368/submissions/57100147

Can anyone provide counter case for my solution in D?

Simply Root the tree at $$$1$$$, We can see for any two nodes to be in minimal tree, we must keep the vertices along their path to their $$$LCA$$$, If we had to add another vertex we must find the Common $$$LCA$$$ of our previous $$$LCA$$$ with the new vertex.

So simply, Find The common $$$LCA$$$ of all $$$K$$$ vertices, then count nodes starting from each needed vertex to the common $$$LCA$$$ by iterating over the parents, and keeping $$$visited$$$ set to avoid double counting, and revisiting the same vertices.

The submission

For E, for a train from u to v first we need to have the x values for all the trains from some station to u. We have a directed graph without any cycles. Use bfs to solve this, add a train to queue only when all the other trains to u have been processed.

Dont know why it is giving TLE

Your code here...
bool dfs(int src,int dest,int sz[],vector<vector<int>>&adj){
     int x = sz[src];
     sz[src]=x+1;
     if(src==dest) return 1;
     for(auto i : adj[src]){ 
         if(!sz[i]&&dfs(i,dest,sz,adj)) return 1;
     }
     sz[src]=x;
     return 0;
}
signed main() { 
    init;
    #ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
    freopen("error.txt", "w", stderr);  
    #endif
    ll _ = 1;
    // cin >> _;
    while (_--) {
     int n,k; cin>>n>>k;
     vector<vector<int>> adj(n+1);
     for(int i=1;i<n;i++){
         int x,y;cin>>x>>y;
         x--;
         y--;
         adj[x].push_back(y);
         adj[y].push_back(x);
     }
     int a[n];
     in(a,n);
     int sz[n]={0};
     for(int i=0;i<k;i++){
         dfs(a[0]-1,a[i]-1,sz,adj);     
     } 
     int cnt=0;
     for(int i=0;i<n;i++){
        cnt+=(sz[i]>0);
     }     
     cout<<cnt<<endl;
  }  
}

Solution for problem D [Minimum Steiner Tree] why this solution giving me WA

Does someone use Systems of Difference Constraints to solve Problem E?

My submission