120 - Stacks of Flapjacks

[ishtiaq ahmed]

removed the code

[newton]

dear ishtiaq

you have made your programe so complex
try to make it simple

try these input:

Code: Select all

              1 9 33 4 45 2 78 100

output should be:

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              1 9 33 4 45 2 78 100
              4 3 5 4 6 0 

but your programe shows

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              1 9 33 4 45 2 78 100
              7 2 3 5 4 7 0

now try to sholve it
best of luck




newton................. simply the best!

[newton]

dear imLazy,

you should delete your code afte AC.










newton ..................... simply the best player.

[joehuang92]

I've tried some easy test cases on papers, then I found some strange situations...

The most popular algorithm discussed in previous threads is to find the largest diameter pancake and flip it to the top, then reversed the stack. Without doubt this algorithm works well to any cases.

But I found that this algorythm doesn't give out the shortest path sorting the stack. Here's an example.

1 3 5 4 2

Followed by the algorithm, the sorting process should be like this:

1 3 5 4 2 flip(3)
5 3 1 4 2 flip(1)
2 4 1 3 5 flip(4)
4 2 1 3 5 flip(2)
3 1 2 4 5 flip(3)
2 1 3 4 5 flip(2)
1 2 3 4 5 flip(0)...done

And it takes 6 flips to complete sorting the stack.

However if I tried another path:

1 3 5 4 2 flip(2)
4 5 3 1 2 flip(4)
5 4 3 1 2 flip(1)
2 1 3 4 5 flip(4)
1 2 3 4 5 flip(0)...done

It only takes 4 flips and the stack is sorted.

Then I rechecked the problem, only to find that it didn't put any limits to the output. Does it means that all the paths, short or long, can be accepted? Or does it require a best solution to get AC?

[joehuang92]

Well, to my test using the algorithm discussed above, I get AC. So the output doesn't need to be the shortest path.

[nbauernf]

To much surprise, this problem is actually somewhat prominent in Academia. In fact, the minimum number of flips for all n! orderings of n pancakes is unknown for n >= 14. It cannot exhaustively be searched because 14! (factorial) is 87,178,291,200. There are many papers in academia bounding the number of flips to go from any pancake stack to any other ordering of the pancakes.

Another fun fact is that the only paper Bill Gates ever published created a tighter bound on the maximum number of flips necessary for any given n.

And so one could turn this into an A* problem with the heuristic of the minimum number of flips required (can be counted by seeing that you need to stick the spatula between any two pancakes that do not belong next to each other in the final ordering). Fortunately, however, this is not necessary =).

[owokko]

I get a problem with 120. Can anyone help me.

I send my code to Online Judge and get "Runtime Error (Signal 11) "

Is any function I did not include??

here are my code :

Code: Select all

//Question http://acm.uva.es/p/v1/120.html
//CPU:  0.002, Memory: Minimum
//state: Runtime Error (Signal 11)  

#include <iostream> 
#include <cstring>

using namespace std;

void sortPancake(int *r, int n)     //array R have 5 element 
{
    for(int j=1; j<n; j++)  //Insertion Sort 
	{
		int k = r[j];
		int i = j-1;
		while(i>=0 && r[i]>k){	
		    r[i+1] = r[i];
            i--;
        }
		r[i+1] = k;
    }           		   
}

void flip(int *r, int n)
{
    for(int i=0; i<=n/2; i++)
    {
        int temp = r[i];
        r[i] = r[n-i];
        r[n-i] = temp;    
    }
} 
  
int main()
{  
    int p[30]={0}, count=0, r[30]={0}, input;
    
    while(cin >> input)
    { 
        p[count++] = r[count] = input;   
          if(cin.eof())
            break; 
    }
    
    for(int i=0; i<count; i++)
        cout << p[i] << " ";
    cout << "\n"; 
        
    sortPancake(r, count);  //Insertion Sort 
           
    for(int i=count-1; i>=0; i--)
    {
        if(p[i]==r[i]) 
            continue;
         
        if(p[0]!=r[i])
        {
            int position = 0;
        
            while(p[position]!=r[i])
                position++;
           
            flip(p, position);
            cout << count-position << " ";
        }
            
        flip(p, i);             //p[0]==r[i] 
        cout << count-i << " ";            
    }
    
    cout << "0"; 
    
    //system("pause");     
    return (0);
}

And other question is "The input is terminated by end-of-file" !?

I use

Code: Select all

  if(cin.eof())    break; 

Is it correct?

[Jan]

Search the board first. Dont open a new thread if there is one already. And for your question the answer is 'no'.

[algoJo]

I got multiple WA on this problem.
Can someone check the correctness of this I/O set?

INPUT

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     2
 33  21    45
33 55
1
2 1
1 2
12 45 1 2 4 7 5 11 21 99 12 47

56 12 48
75 2 1 44 53 84 21 46 82 43 99 49 89 7 9
 1 2 3 6 7 8 4 5
1 3  19 4 95 62 42 43 13 2
1 2 3 4 5 6 7 8 9 10 11 12 14 13 15 16 17 18 19 20

OUTPUT

Code: Select all

2
0
33 21 45
1 1 0
33 55
0
1
0
2 1
1 0
1 2
0
12 45 1 2 4 7 5 11 21 99 12 47
3 1 2 11 3 11 4 6 6 7 7 11 8 0
56 12 48
1 2 0
75 2 1 44 53 84 21 46 82 43 99 49 89 7 9
5 1 13 2 10 3 11 4 5 9 6 11 7 14 8 13 9 12 10 13 11 12 13 13 0
1 2 3 6 7 8 4 5
3 1 6 2 5 3 7 4 0
1 3 19 4 95 62 42 43 13 2
6 1 6 2 4 3 4 8 5 6 8 7 9 8 0
1 2 3 4 5 6 7 8 9 10 11 12 14 13 15 16 17 18 19 20
1 1 2 3 3 4 5 5 6 8 7 8 0

Can someone post more tricky I/O set?
Thanks

[anderson101866]

I have seen several topics about Q120.

I tested several test case and I got correct output.

But I still got WA...

Could someone who got AC give me more test case?

thank you!

[Jan]

Dont open a new thread if there is one already.

[mohsincsedu]

To Sedefcho:

Plz correct ur i/o set....



To algoJo:
The problem statement:

all pancakes separated by a space.

So, there is no such Input that u given...

[anderson101866]

INPUT:

Code: Select all

...
4 4 1
4 4 5 4 4 1
4 4 5 4
99 7 99 0 7 99 0
2 4 4 4 2 4 
11 13 15 17 13 15 11
20 20 200 20 20 200 20 20

I think you maybe forgot one condition in the problem.
"All pancakes in a stack have different diameters," so there are no two things of the same size. I post some test cases again.
May it be helpful~
INPUT

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8 4 6 7 5 2
1 2 3 5 4
7 3 5 4 8 9 10 1 13
2
33 21 45
33 55
1
2 1
1 2
56 12 48
75 2 1 44 53 84 21 46 82 43 99 49 89 7 9
1 2 3 6 7 8 4 5
1 3 19 4 95 62 42 43 13 2
1 2 3 4 5 6 7 8 9 10 11 12 14 13 15 16 17 18 19 20

OUTPUT

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8 4 6 7 5 2
1 4 2 5 3 4 5 0
1 2 3 5 4
2 1 2 3 0
7 3 5 4 8 9 10 1 13
3 2 8 5 8 6 0
2
0
33 21 45
2 0
33 55
0
1
0
2 1
1 0
1 2
0
56 12 48
1 2 0
75 2 1 44 53 84 21 46 82 43 99 49 89 7 9
5 1 13 2 11 3 6 4 10 5 14 6 7 10 8 12 9 11 14 12 14 0
1 2 3 6 7 8 4 5
3 1 4 6 0
1 3 19 4 95 62 42 43 13 2
6 1 6 2 4 3 4 6 5 9 6 8 9 0
1 2 3 4 5 6 7 8 9 10 11 12 14 13 15 16 17 18 19 20
8 7 8 9 0

[amacfie]

be aware that the bring-to-top method is not always optimal:

http://www.google.com/url?sa=t&source=w ... CfG0QBaWIg

[vinocit]

Could any one temme wats wrong with my prog

Code: Select all

#include<iostream>
#include<sstream>
#include<stack>
using namespace std;
int min(int a[],int i,int n)
{
	int m=i;
	for(int j=i;j<n;j++)
		if(a[m]>a[j])m=j;
	return m;
}
void flip(int a[],int ind,int n)
{
	stack<int> sta;
	for(int i=ind;i<n;i++)
		sta.push(a[i]);
	for(int i=ind;i<n;i++)
	{
		a[i]=sta.top();
		sta.pop();
	}
}
int main()
{
	int sta[100];
	char s[300];
	while(cin.getline(s,300))
	{
		stringstream sin(s);
		int i=0;
		while(sin>>sta[i++])cout<<sta[i-1]<<" ";
		int n=i-1;
		cout<<"\n";
		for(i=0;i<n-1;i++)
		{
			int ind=min(sta,i,n);
			if(ind!=i)
			{
				if(ind<n-1)
				{
					cout<<ind+1<<" ";
					flip(sta,ind,n);
				}
				cout<<i+1<<" ";
				flip(sta,i,n);
			}
		}
		cout<<"0\n";
	}
	return 0;
}

If there are more than one solution is there any criteria to select one among them? :|