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Codeforces Round #185 Editorial
By lydrainbowcat, 11 years ago, In English

311A - Ближайшая пара

http://codeforces.net/blog/entry/7787

311B - Транспортировка кошек

P.S. I feel very sorry that I thought it was a traditional DP problem with only 800B code and didn't realize some participants were not familiar with such kind of problems, so I said it was easy.

Let a[i] be the distance from hill 1 to hill i, s[i]=a[1]+a[2]+…+a[i].

Firstly, we sort the cats by (Ti-a[i]). Then we can divide the cats into P consecutive parts, and plan a feeder for each part. Dynamic Programming can solve this problem.

Let f[i,j] indicates the minimum sum of waiting time with i feeders and j cats.

f[i,j] = f[i-1,k]+a[j]*(j-k)-s[j]+s[k] = a[j]*j-s[j] + f[i-1,k]+s[k]-a[j]*k

That’s O(PM^2). It’ll get TLE.

Let p>q, if p is "better" than q, then:

f[i-1,p]+s[p]-a[j]*p>f[i-1,q]+s[q]-a[j]*q

(f[i-1,p]+s[p])-(f[i-1,q]+s[q])>a[j]*(p-q)

g[p]-g[q]>a[j]*(p-q)

So we can use Convex hull trick with a queue. Then we get O(MP), which can pass the problem.

311C - Достать сокровище

Firstly, we solve such a problem: if we can go exactly k,k1,k2……or kp cells forward each step, what cells can we reach?

We divide the H cells into k groups: Group 0,1……k-1. The i-th cell should be in Group (i mod k).

  • If we reach Cell x in Group (x mod k), we can also reach Cell (x+kj , 1<=j<=p) in Group ((x+kj)mod k).
  • If we reach Cell x in Group (x mod k), we can also reach Cell (x+k) in the same group.

Let D[i] be the minimum cell we can reach in Group i. Then we can reach all the cells which number are bigger then D[i] in Group i.

Regard the groups as points. Regard k,k1,k2……kp as edges. And use a Shortest-Path Algorithm to calculate all D[i].

Notice that there are at most 20 operations of type 1, we are able to run such an algorithm after each of these operations. The total time complexity is O(20*k*20*log(k)) with Dijkstra.

Secondly, we build a binary-heap to solve operations of type 2 and 3.

  • Type1 – If a D[i] becomes smaller, we put the new cells we can reach into the heap.
  • Type2 – Decrease a value of a node in the heap, or a cell in the “Treasure Cell” array.
  • Type3 – Ask the biggest node in the heap and delete it.

These are basic operations of binary-heap. The time complexity is O(NlogN). In C++, you can also use priority_queue from STL and lazy-tags. So we can solve the whole problem in O(400klogk+NlogN).

311D - Куб на отрезке

http://codeforces.net/blog/entry/7787

311E - Биолог

http://codeforces.net/blog/entry/7847

312A - Чье это предложение?

We only need to find out if “miao.” is a prefix of the sentence, and if “lala.” is a suffix of the sentence. Pay attention to the situation when both conditions are satisfied.

312B - Лучники

http://codeforces.net/blog/entry/7847

  • Vote: I like it
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Comments Comments (18)
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11 years ago, # |
  Vote: I like it +1 Vote: I do not like it

What is slope optimization?

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11 years ago, # |
  Vote: I like it +7 Vote: I do not like it

I'm sorry, but I dont understand why "f[i,j] = f[i-1,k]+a[j]*(j-k)-s[j]+s[k] = a[j]*j-s[j] + f[i-1,k]+s[k]-a[j]*k" Can anybody explain me?

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    11 years ago, # ^ |
      Vote: I like it +8 Vote: I do not like it

    It's probably this:

    f[i,j] = min{f[i-1,k]+a[j]*(j-k)-s[j]+s[k] | k < j}

    This "a[j]*(j-k)-s[j]+s[k]" calculates the total waiting time for those cats between k and j, assuming the (i-1)-th feeder takes care of the k-th cat and the i-th feeder takes care of cats between k and j.

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      11 years ago, # ^ |
        Vote: I like it +3 Vote: I do not like it

      What I found confusing is the usage of a[j], in the above expression, where j represents j cats, whereas a[i] was originally stated to be "Let a[i] the distance from hill 1 to hill i". Also, the usage of "s[i] = a[1]+a[2]+…+a[i]." is not quite clear with the definition of a[i]. What does it represent? Any clarification will be greatly appreciated.

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        11 years ago, # ^ |
        ← Rev. 3   Vote: I like it +20 Vote: I do not like it

        Consider the array that contains values p[i], the prefix sum of height values where p[i] = h[1] + h[2] +... + h[i]. Now consider array
        a[i] = T[i] — p[c[i]], where c[i] denotes the hill number ith cat goes to. Sort this array. Now f(i, j) denotes the minimum time taken by i feeders to pick up the first j cats from the array 'a'. Recursion can be written as
        f(i, j) = times taken by i-1 feeders to pick up first k cats + time taken by ith feeder to pick up cats from k+1 to j. This can be written as
        f(i, j) = min{f(i-1, k) + (j-k)*a[j] — s[j] + s[k]} 1 <= k <= j
        As for the s[k] — s[j], you can interpret it like this. Each -a[i] denotes how much time the cat would have to wait, if a feeder starts at time 0. If feeder starts at time x, it will experience a delay of x — a[i]. For cats from k+1 to j, the feeder starts from his position at a[j] (you have to convince yourself this is correct). So the delay experienced by cats from k+1 to j is (a[j] — a[k+1] + a[j] — a[k+2] + .... a[j] — a[j]) = (j-k)*a[j] — (s[j] — s[k]) Correct me if I am wrong.

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          11 years ago, # ^ |
            Vote: I like it 0 Vote: I do not like it

          Thanks you, now its clear for me.

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          11 years ago, # ^ |
          ← Rev. 2   Vote: I like it 0 Vote: I do not like it

          Great explanation. Thanks very much. Just one thing: I suppose it should be 0 <= k < j.

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          11 years ago, # ^ |
            Vote: I like it +11 Vote: I do not like it

          Thanks for such a detailed and articulate explanation. I wish the editorials were written with a similar clarity and rigor.

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11 years ago, # |
← Rev. 2   Vote: I like it +2 Vote: I do not like it

Hi, I can not understand why the formula given for f[i, j] is correct. Can somebody explain it to me? Thanks

Edit: In the contest I got another formula(suppose b[i] = T[i] — a[h[i]]): f[i — 1, k] + b[i] * (j — k) But I can't understand why these two are the same?

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11 years ago, # |
  Vote: I like it 0 Vote: I do not like it

311B / 312D In my view, it will make more sense if we add an additional constraint that all feeders should start at a non-negative time.

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5 years ago, # |
  Vote: I like it 0 Vote: I do not like it

For 311A, I submitted

FOR i from 1 to n
    PRINT (n-i,0)

Here p[j].x — p[i].x is always negative but the checker gave wrong answer on test 4. Is this the same mistake that you've mentioned?

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5 years ago, # |
  Vote: I like it 0 Vote: I do not like it

311B the feeder can actually start at any time including the negatives.

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    5 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Thank you for pointing that out! it was driving me crazy!

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5 years ago, # |
← Rev. 2   Vote: I like it +5 Vote: I do not like it

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