hoco's blog

spoj _ nested dolls
By hoco, 11 years ago, In English

hi everybody, about this problem + if any body solved it, plz share his code.

My algorithm is to use LIS and see how many distinct "Increasing sequences" there is. but I think it's hard to code. does any body have any easier algorithm?

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11 years ago, # |
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Nested Dolls

You're gonna need to take a look here Online Algorithms for Dilworth's Chain Partition and here Dilworth's theorem, and I think you will get it...

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    11 years ago, # ^ |
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    i think your first link has a problem.

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      11 years ago, # ^ |
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      At least for me it works, its a pdf file...

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        11 years ago, # ^ |
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        mine doesn't

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          11 years ago, # ^ |
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          it's working on my machine well too. I uploaded the file in another host ... maybe it'll help you :)

          link

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11 years ago, # |
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http://ncpc.idi.ntnu.no/ncpc2007/ncpc2007solutions.pdf

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    11 years ago, # ^ |
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    The solution in the editorial is easier. O(NlogK), where K is the length of largest antichain.

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      11 years ago, # ^ |
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      Can you please tell me in which editorial I will get it. I am stuck with the same problem.

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        11 years ago, # ^ |
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        Its the one sparik1997 posted. Check the solution of problem G.

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11 years ago, # |
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Find The Longest Decreasing Subsequence But keep in mind that you can take 2 equal values like 3 3 2 2 1 1 is a valid decreasing subsequence

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11 years ago, # |
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It is my solution :
Prerequisite : range tree
MAXN = 1000 highest value of width and height
T[1000] is array and all elemnt of T is zero
query(x, 1, MAXN, 1) = find the biggest number which is small or equal to x
upd(x, y, 1, MAXN, 1) is T[x] = y; 

    scanf("%d",&n);

    for(int h=0; h<n; h++)
    {
       int a, b;
       scanf("%d %d",&a,&b);
       D[a].push_back(b);
    }

    for(int h=1; h<=MAXN; h++)
    {
       for(int j=0; j<(int)D[h].size(); j++)
       {
         int k = query(D[h][j]-1, 1, MAXN, 1);

         L[k]--;

         if(!L[k])    upd(k, 0, 1, MAXN, 1);
       }

       for(int j=0; j<(int)D[h].size(); j++) upd(D[h][j], D[h][j], 1, MAXN, 1),L[D[h][j]]++;
    }

    for(int h=1; h<=MAXN; h++) ans += L[h];

    printf("%d\n",ans);
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10 years ago, # |
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my greedy approach

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