Come on people,why downvote? I made some prediction here and as you can see the 1st problem really had some sign of google translation: "Little Susie, thanks to her older brother, likes to play with cars." Didn't my prediction work?

Many strong coders won't take part due to world finals . All of the strong coders are in div1 . That's why there are many div1 contests consecutively lined up after the WF ends.

now 1 min :D

Well, the author thought that good luck and inspiration are enough :)

Do you believe in EDITORIALS??

Telling what exactly you think is wrong in a translation is a more constructive way rather than just saying you hate someone.

E was so obvious that I thought my logic has got to be wrong.

Guy registered 3 hours ago judges the problemset. You even weren't brave enough to participate from your real account, lol =)

How do you use Prims?

Dijkstra with set doesnt give TL on pretests

This solution may be hacked by following test case:

3 3 1 2 8 1 3 6 2 3 2 1

Simply doing djikstra — it will output 14; correct answer is 8.

Is that always true anyway? At least my implementation of Dijkstra does not have that property, and I spent forever trying to figure out how to modify it to get that property.

its wrong, so you cannot prove it :D
UPD: I wasn't fast enough

it isnt because i used dijkstra and i got WA. i think you should modify dijkstra so when (d[v] == d[u] + w) happens, you change d[v] to d[u]+w if the trees wieght decreases.

Let d[x] denote the minimum distance from u to x. Let's construct a tree in G using this approach: 1. Let u be the root. 2. For all other vertex p, we can choose as a parent any node q such that d[p]=d[q]+w[q][p]. The most natural idea is to select the q with minimum w[q][p].

Since for each vertex we are selecting the edge with minimum weight, the total sum of weights will be minimum. It just remains to prove that for two different vertices, we can't choose the same edge as a link to the parent node (this is a consequence of positive weights)

Actually solution algo was directly google-able too :P one can get ready-made algo for E in 2 clicks :/

It's obvious I think :) But D is easier than C.

Hard ?? why ?? it was a simple implementation .

I am From Nepal (Original) . Please Don't Make a Joke about Earthquake

I am from north India. There was no such earthquake.

if the difficulty level was "standard" ANY of problems A,B,C,D could be Div2. A or at most Div2. B.Because no specific algos were needed,and problem C had some tricky coroners and wasn't actually difficult

You will get rating after system tests and rating recalculation

It was DP. You should initialize 0th tree at -infinity(-10^14) and n+1th tree at infinity.

What's the problem?

Have you actually participated in those, because I certainly feel that overall level of lunchtimes is at par with Div2 contests(old ones when I used to participate :)) and most of the times even difficult than that.

IOI is also for school students. Do you think it's easy?

There have been many Java AC too ,so probably your implementation could have been slightly slow ?

Because we sort the array first, so both of them will produce 1 1 1 1 1 1 1 5. The answer should be three: serve the first two, and serve the last person.

Hope it helps!

o_O what a grammar ?!

The same situation is in my room, 1014.

Code doesn't look 100% similar. Their outputs on the third test differ also.

This is Common Problem With Short code. Two code can Have similar Idea ( Even These Code seems to be same but not Exactly Same). And another thing is, Generally Student from same University have closer Idea about many problems.

System testing wasn't finished. It is finished now. Your rating's 1448, gratz on your first contest!

Congrats

When a[i][j] == 3 || a[i][j] == 1 you should skip entry. i cant understand your code seems like you do the same for a[i][j] == 2 and it give WA.

A A B A D in implementation imho

its simple because you assume tree number 2 will always be cut in code: int ans=1+func(2,1); ans=max(ans,1+func(2,2));