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### 473 - Raucous Rockers

Posted: Fri Feb 04, 2005 5:31 pm
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

Posted: Fri Feb 04, 2005 8:58 pm
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.

Posted: Fri Feb 04, 2005 9:59 pm
I also just found a counterexample for your current greedy algorithm:
4 11 2
9, 10, 2, 11

optimal solution is:
9 2
11
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.

Posted: Sat Feb 05, 2005 10:41 pm
Thanks, I just got AC with an O(n*m*t) DP algorithm.

Posted: Sun May 08, 2005 10:01 pm

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?

Posted: Mon May 09, 2005 10:00 am
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.

Posted: Tue May 10, 2005 8:36 pm
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()).

### Can you tell me what the ouputs are of these inputs?(ACM473)

Posted: Fri Aug 04, 2006 2:06 pm
I got WA on this problem

Now I want to find out why it was wrong

But I need the ouputs of these inputs

Can anyone help me? Thanks

Code: Select all

``````3

8 4 3
2, 4, 8, 1, 1, 2, 3, 5

7 4 2
3, 4, 4, 1, 4, 4, 4

10 5 7
3, 6, 9, 1, 9, 1, 5, 10, 10, 2
``````

### 473...WA,need I/O

Posted: Thu Sep 07, 2006 12:29 pm
And can anybody tell me what is the upper bound of the number n,t and m?

### Re: 473...WA,need I/O

Posted: Tue Nov 04, 2008 12:40 pm
You can check this post (http://acm.uva.es/board/viewtopic.php?f ... 82dea1139a), and the upper bound of n, t, and m are clearly illustrated in there.

### Re: 473 - Raucous Rockers

Posted: Sun Feb 28, 2010 4:08 pm
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?

### Re: 473 - Raucous Rockers

Posted: Wed May 05, 2010 9:34 am
its easy to solve in N^2*m..but i can't get he m*n*t solution
Could anyone Give some hints ....?

### 473 - Raucous Rockers - RTE

Posted: Fri Mar 11, 2011 4:40 pm
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 parĂ¢metros
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;
}
}
``````

### Re: 473 - Raucous Rockers

Posted: Mon Jul 04, 2011 9:17 pm
n<=800 , m<=100 and t<=100.
I got AC with this limits.

### Re: 473 - Raucous Rockers

Posted: Fri Jul 27, 2012 2:16 am
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;
}
``````