Thanks! I'm trying my best to make all that happen. ;)

Meanwhile me "forever-single"

Don't worry, you'll get to help him in the contest. XD

Sorry about that. I've informed the coordinators of this. :)

EDIT: Len and abhishek_saini, it has been fixed.

Thanks! Hope you'll like my problems too.

I am unable to see the predictions.

Thank you as always ! :D

lets just stay up late tonight

lets just stay up late tonight

lets just stay up late tonight

same T_T

*opportunity

It looks like : A-A-B-C-D-E :)

I think "NO" !!

...You made me realize that i did the same thing lol

Damn it! I did the same thing.

Same to me

same here, read it only 15 minutes before the end :\

rating army Lmao

And I'm pretty sure code obfuscation is against the rules!

Funny that one of these bots was hacked: http://codeforces.com/contest/796/submission/26272062

Hello guy, I wanna know how to show predicted rating changes in standing page like yours, some extensions in browser? Thank you!

My solution works in O((n + m)logn). Please stay tuned for the editorial.

I guess because the tree is arranged according to the law. So the question is actually how to destroy some roads in order to get number_of_cities_with_police connected components. The best way is to have one police station in every connected component. And because in the tree every city is at most d units away, it is possible to keep this property in partition at the end. I hope you understand :)

Now I understand why my solution in D was wrong :(

In problem D we can run a single bfs from every police station. Then for every city we have its distance to its closest police station as well as the number of different police stations that has its distance equal to the shortest distance. Run a single dfs from vertex 1. For each vertex, let's consider its children. If a children has abs(dis[v]-dis[u])!=1 then we will cut that edge. If dis[v] == dis[u] — 1 then we will count how many vertex like that and we will erase (cnt-1) of them because we only need one of them to maintain the distance. And for (dis[v] == dis[u] + 1), as we already have the number of ways to get to this vertex, we will delete this edge if way[v] > 1 and then way[v]--. That's all!

D is a simultaneously BFS from every city with police. Mark each connected city with nubmer of a source police city, then go through edges and if a "police number" on the left != one on the right, then the edge can be removed. You even don't need the d parameter.

2

My program prints 2.

2 1 1 1 1 2

i hack using this : 2 1 1 1 1 2

2 1 1 1 1 2

the answer is 1

TL hack with 10^6 numbers. OK if solution hasn't scanf/sync_with_stdio.

(wrong solution)

I think starting from guard with a max strength and connected to maximum number of other max guards is correct, your algorithm will not work for graph: 10 10 10 10 1 1 1 1 ... 1 2 1 3 1 4 4 5 4 6 4 7 ...

Your algorithm starts from 4, and we should start from 1.

Each Id belongs to one person I guess! but if each code are same then they will be skipped without the first one. But he should get punished -_-

I think we can separate the tree in forest, where the root of each tree has a police office in it. This way the answer will always be P - 1

I don't think it is. I was failing on 6 for the longest time too. What does your solution output for this inout:

5 2 2
1 4
1 2
2 3
4 3
3 5

I believe the correct output is

1
2

(EDIT: Fixed it from 5 2 3 to 5 2 2)

Damn. I failed to notice that too. My program was based on the assumption that there's only one police in one city ;-;

Let's see.

You've got to be kidding me...

The reason why system test is pending

1 ≤ h_i ≤ 10⁶, so you can keep them in bool array, like holes[hole] = true.

Then check if holes[bone] = true then print bone, break else continue swapping.

Link for submission: http://ideone.com/BvFVqI

1) if first position contains hole , then obviously print 1 , return 0; 2) else -------> keep track of current position of bone in any variable(say boneVar) -------> Initial value of boneVal = 1(given in question)
----- > now iterate for k times and take values of cups to be swapped,,, suppose inputs are cup1 , cup2 for(int i = 0 ; i < k ; i++)

for(int i = 0 ; i < k ; i++)
{ 
         cin >> cup1 >> cup2;
          if(cup1 == boneVar) boneVar = cup2;
          else if(cup2 == boneVar) boneVar = cup1;
          /// now check if the new position of bone contains hole then that is answer
         if(hole[boneVar])
         {
              cout << boneVar;return 0;
         }
}

if(cup1 == boneVar) boneVar = cup2;
      else if(cup2 == boneVar) boneVar = cup1;

I think it may take a while because as you can see, there are many many spam accounts in today round and moreover, their owners have used "Find and Replace" to replace names of all variables and functions which will make the CF anti-cheat accept their solutions UPD: All bots have been deleted

If (p / 2) * k >= n. Obviously, It is the maximal point. Else we use dp with n * p * k states, each state has O(k) transitions.

koil (save the name to remove )

Not for all...:( :(

Did it pass ???

Yep, my code TLEd on that test aswell http://codeforces.com/contest/796/submission/26262616

Can we have an rejudge since my code is only reading the input in C++?

Why we should get TLE only for cin?? Even in CF ??
This should be rejudged :'(

slow io

Big input.... Feel ya

I know right? I used multiset and got TLE... if TL was 3000 it would have passed. I'm pretty sure they are not gonna rejudge though, unfortunately :/

I use multiset (C++) and it passed:)

After this comment I've checked time of my solution ... it is 1949!

My solution with Segment tree passed in 1123 ms.

My solution using multiset passed with 1.85

Strange... I got AC with 1231 ms http://codeforces.com/contest/796/submission/26271625