POJ 3126, Prime Path

Time Limit: 1000MS  Memory Limit: 65536K
Total Submissions: 3134  Accepted: 1911

Description
The ministers of the cabinet were quite upset by the message from the Chief of Security stating that they would all have to change the four-digit room numbers on their offices.
— It is a matter of security to change such things every now and then, to keep the enemy in the dark.
— But look, I have chosen my number 1033 for good reasons. I am the Prime minister, you know!
— I know, so therefore your new number 8179 is also a prime. You will just have to paste four new digits over the four old ones on your office door.
— No, it’s not that simple. Suppose that I change the first digit to an 8, then the number will read 8033 which is not a prime!
— I see, being the prime minister you cannot stand having a non-prime number on your door even for a few seconds.
— Correct! So I must invent a scheme for going from 1033 to 8179 by a path of prime numbers where only one digit is changed from one prime to the next prime.

Now, the minister of finance, who had been eavesdropping, intervened.
— No unnecessary expenditure, please! I happen to know that the price of a digit is one pound.
— Hmm, in that case I need a computer program to minimize the cost. You don't know some very cheap software gurus, do you?
— In fact, I do. You see, there is this programming contest going on... Help the prime minister to find the cheapest prime path between any two given four-digit primes! The first digit must be nonzero, of course. Here is a solution in the case above.
1033
1733
3733
3739
3779
8779
8179
The cost of this solution is 6 pounds. Note that the digit 1 which got pasted over in step 2 can not be reused in the last step – a new 1 must be purchased.

 

Input
One line with a positive number: the number of test cases (at most 100). Then for each test case, one line with two numbers separated by a blank. Both numbers are four-digit primes (without leading zeros).

 

Output
One line for each case, either with a number stating the minimal cost or containing the word Impossible.

 

Sample Input
3
1033 8179
1373 8017
1033 1033

 

Sample Output
6
7
0

 

Source
Northwestern Europe 2006

---

//  POJ3126.cpp : Defines the entry point for the console application.
//

#include  < iostream >
#include  < queue >
using   namespace  std;

int  main( int  argc,  char *  argv[])
{
     // init prime table
     bool  prime[ 10001 ];
    memset(prime, 1 , sizeof  (prime));
    prime[ 0 ]  =   false ;prime[ 1 ]  =   false ;
     for  ( int  i  =   2 ; i  <   5000 ;  ++ i)
    {
         int  k  =   2 ;
         int  next;
         while ((next  =  (k ++ )  *  i)  <   10000 )
            prime[next]  =   false ;
    }

     int  cases;
    cin  >>  cases;
     int  beg, end;
     int  step;
     int  visited[ 10001 ];
     int   base [ 4 ] =  { 1 ,  10 ,  100 ,  1000 };
     for  ( int  c  =   0 ; c  <  cases;  ++ c)
    {
        scanf( " %d %d " ,  & beg,  & end);
        memset(visited,  - 1 ,  sizeof (visited));
        queue < int >  q;
        q.push(beg);
        visited[beg]  =   0 ;
        step  =   - 1 ;

         while  ( ! q.empty())
        {
             int  cur  =  q.front();
            q.pop();
             if  (cur  ==  end)
            {
                step  =  visited[cur];
                 break ;
            }
            
             for  ( int  i  =   0 ; i  <   4 ;  ++ i)
            {
                 for  ( int  j  =   0 ; j  <   10 ;  ++ j)
                {
                     int  l  =  cur  /  ( base [i]  *   10 );
                     int  r  =  cur  %   base [i];
                     int  num  =  l  *   base [i]  *   10   +  j  *   base [i]  +  r;
                     if  (num  >   1000   &&  num  !=  cur  &&  prime[num]  ==   true   &&  visited[num]  ==   - 1 )
                    {
                        q.push(num);
                        visited[num]  =  visited[cur]  +   1 ;
                    }
                }
            }
        }

         if  (step  ==   - 1 )cout  <<   " Impossible\n " ;
         else  cout  <<  step  <<  endl;
    }

     return   0 ;
}