473 - Raucous Rockers

[Christian Schuster]

Hello,

I tried to solve problem 473 by a greedy algorithm:

1. S := sequence of song lengths (as given in input)
2. determine number of disks needed for the songs in S
3. if disks <= m return |S|
4. otherwise remove longest song from S and go to 2.

I assumed that the song order given in the input must be maintained over all disks, so I put songs on disk 1 until it's full, then on disk 2 until it's full, and so on.

I didn't find any case where the result differs from my (too slow) backtracking solution. Obviously, there must be such cases, because I received WA.

Thanks,
Christian

[Adrian Kuegel]

Your greedy is correct if you apply it to one disk only. So to get a correct overall algorithm, use dynamic programming to partition the n songs into m intervals of songs which will be assigned to disks. Within one disk, apply your greedy algorithm until the songs fit on that disk. The dynamic programming part is similar to problems 709 and 10239.

[Adrian Kuegel]

I also just found a counterexample for your current greedy algorithm:
4 11 2
9, 10, 2, 11

optimal solution is:
9 2
11
but your greedy will get:
9
2

unfortunately, the O(n^2 * m) dynamic programming algorithm I have is too slow. And the pruning I inserted seems to be wrong.

[Christian Schuster]

Thanks, I just got AC with an O(n*m*t) DP algorithm.

[Abednego]

Code: Select all

 The songs will be recorded on the set of disks in the order of the dates they were written.


Does this mean that we will record disk 1 first, then disk 2, then 3, etc.? Or can we record song 1 onto disk 1, then song 2 onto disk 2 and then song 3 onto disk 1 again?

[Christian Schuster]

It's not allowed to put songs 1 and 3 on disk 1 and song 2 on disk 2. More formally, for every possible pair (a,b) of chosen song numbers having a>b, the relation disk(a)>=disk(b) must hold.

[Abednego]

Ok. I got it accepted in 2 seconds with a O(n*m*t) DP using O(m*t) memory. n is at most 10240, t is at most 128 and m is at most 128. They are probably smaller than this, but they are certainly not larger (checked with assert()).

[rnd]

Hello,
Does this problem is a binary knapsack problem with some constraint? The recursion formula:

Code: Select all

M(i, j) = M(i-1, j) if j-t(i) and j belongs to diffrent discs
M(i, j) = max{M(i-1, j), M(i-1, j-t(i)) + 1} otherwise
1 <= i <= n
1<= j <= m*t
M(i, j) - maximum number of songs from range 1..i that could be placed on discs with whole size of j

The second line is standard 1/0 knapsack formula. The first line gives constraint that single track cannot begin in first CD and ens at second?
Does my reasoning is correct?

[Angeh]

its easy to solve in N^2*m..but i can't get he m*n*t solution
Could anyone Give some hints ....?

thanks in advance ...

[fernandohbc]

This code works on my box, but keeps giving me RTEs.
Any ideas?

Code: Select all

#include <iostream>

#define MAX 2000000

using namespace std;

struct Disk {
  int totalTime;
  int capacity;
};

class DiskCollection {
   private:
       Disk currentDisk;
       int totalTracks;
       int diskToWrite;
       int qtdDisks;

  public:
  DiskCollection(int qtdDisks, int capacity) {
      this->currentDisk.totalTime = 0;
      this->currentDisk.capacity = capacity;
      this->qtdDisks = qtdDisks;
      this->totalTracks = 0;
      this->diskToWrite = 0;
  }

  int getTotalTracks() {
      return this->totalTracks;
  }

  bool isBetterThan(DiskCollection * diskCollection) {
      if (this->totalTracks > diskCollection->totalTracks) {
          return true;
      }

      if (this->totalTracks == diskCollection->totalTracks) {
          if (this->diskToWrite < diskCollection->diskToWrite) {
              return true;
          }

          if (this->diskToWrite == diskCollection->diskToWrite) {
              if (this->currentDisk.totalTime <
diskCollection->currentDisk.totalTime) {
                  return true;
              }
              return false;
          }
          return false;
      }
      return false;
  }

  DiskCollection * insertTrack(int trackTime) {
      DiskCollection * result = NULL;
      if (this->currentDisk.capacity - this->currentDisk.totalTime >=
trackTime) {
          result = new DiskCollection(this->qtdDisks,
                  this->currentDisk.capacity);
          result->totalTracks = this->totalTracks + 1;
          result->diskToWrite = this->diskToWrite;
          result->currentDisk.capacity = this->currentDisk.capacity;
          result->currentDisk.totalTime = this->currentDisk.totalTime
                  + trackTime;
      } else if (this->diskToWrite != this->qtdDisks - 1) {
          result = new DiskCollection(this->qtdDisks,
                  this->currentDisk.capacity);
          result->totalTracks = this->totalTracks + 1;
          result->diskToWrite = this->diskToWrite + 1;
          result->currentDisk.capacity = this->currentDisk.capacity;
          result->currentDisk.totalTime = trackTime;
      }
      return result;
  }
};

DiskCollection * emptyCollection(int qtdDisks, int capacity) {
 DiskCollection * empty = new DiskCollection(qtdDisks, capacity);
 return empty;
}

int main() {
   char line[MAX];
   cin.getline(line, MAX);
   int qtdTc = atoi(line);

   for (int tc = 1; tc <= qtdTc; tc++) {
       // Ignora linha em branco
       cin.getline(line, MAX);

       // Leitura dos parâmetros
       cin.getline(line, MAX);
       int n = atoi(strtok(line, " "));
       int t = atoi(strtok(NULL, " "));
       int m = atoi(strtok(NULL, " "));

       // Ignora o resto da linha
       cin.getline(line, MAX);
       char * tok = strtok(line, " ,");
       int trackTimes[n + 1];
       int i = 1;
       while (tok) {
           trackTimes[i++] = atoi(tok);
           tok = strtok(NULL, " ,");
       }

       DiskCollection * dc[n + 1][t * m + 1];
       for (i = 0; i <= n; i++) {
           for (int j = 0; j <= t * m; j++) {
               if (i == 0) {
                   dc[i][j] = emptyCollection(m, t);
               } else if (j - trackTimes[i] < 0) {
                   dc[i][j] = dc[i - 1][j];
               } else {
                   DiskCollection * newCollection = dc[i - 1][j
                              - trackTimes[i]]->insertTrack(trackTimes[i]);
                   if (newCollection
                             && newCollection->isBetterThan(dc[i - 1][j])) {
                       dc[i][j] = newCollection;
                   } else {
                       dc[i][j] = dc[i - 1][j];
                   }
               }
           }
       }
       cout << dc[n][t * m]->getTotalTracks() << endl << endl;
   }
}


[shibly]

n<=800 , m<=100 and t<=100.
I got AC with this limits.

[shopnobaj_raju]

Got WA . Can anoyone help??

Code: Select all

#include<cstdio>
#include<iostream>
#include<vector>
#include<utility>
#include<algorithm>
using namespace std;
int main()
{
    int c,i,j,k,ans,n,t,m,temp,tem2,tk,x;
    vector<int> songs;
    vector<vector<int> > dp;
    scanf("%d",&c);
    for(x=1;x<=c;x++)
    {
        songs.clear();
        dp.clear();
        scanf("%d %d %d",&n,&t,&m);
        for(i=0;i<n-1;i++)
        {
            scanf("%d, ",&temp);
            songs.push_back(temp);

        }
        scanf("%d",&temp);
        songs.push_back(temp);
        //for(i=0;i<n;i++) cout<<songs[i]<<' ';
        dp.resize(n+1);
        for(i=0;i<dp.size();i++) dp[i].resize(m+1);
        for(i=0;i<=m;i++) dp[n][i]=0;
        for(i=0;i<=n;i++) dp[i][0]=0;
        if(songs[n-1]<=t) for(i=1;i<=m;i++) dp[n-1][i]=1;
        else for(i=1;i<=m;i++) dp[n-1][i]=0;
        for(i=n-2;i>=0;i--)
        {
            for(j=1;j<=m;j++)
            {
                dp[i][j]=dp[i+1][j];
                if(songs[i]<=t)
                {
                    temp=0;
                    tk=0;
                    for(k=i;k<n;k++)
                    {
                        if(temp+songs[k]<=t)
                        {
                            tk++;
                            temp+=songs[k];
                            tem2=dp[k+1][j-1];
                            if(tem2+tk>dp[i][j]) dp[i][j]=tem2+tk;
                        }
                    }
                }
            }
        }

        if(x!=1) printf("\n%d\n",dp[0][m]);
        else printf("%d\n",dp[0][m]);
    }
    return 0;
}