poj 3126 广搜 Prime Path

poj 3126 广搜 Prime Path

#include < iostream >
#include < queue >
#include < math.h >
using   namespace  std;
const   int  MAX = 10000 ;
int  visited[ 10000 ];
int  result[ 10000 ];
bool  isprimer( int  npp)
 {
      for  ( int  q  =   2 ; q  <=  sqrt( double (npp));  ++ q)
     {
          if  (npp  %  q  ==   0 )
         {
              return   0 ;
         }
     }
      return   1 ;
 
 }

int  BFS( int  start, int  end) 
{
    queue < int >  q;
     if  (start  ==  end)
         return   0 ;
    q.push(start);
    visited[start] = 1 ;
    result[start] = 0 ;
     while ( ! q.empty()) 
    {
         int  temp = q.front();
        q.pop();
         int  next,j,k;
         int  num[ 3 ];
        
         for  ( int  i  = 0 ;i < 4 ; ++ i)
            {    
                 num[ 0 ] = temp % 10 ;
                 num[ 1 ] = (temp % 100 ) / 10 ;
                 num[ 2 ] = (temp % 1000 ) / 100 ;
                 num[ 3 ] = temp / 1000 ;   
                  for (j = 0 ;j <= 9 ;j ++ )      
               {                       
                 num[i] = j;
                 next = num[ 0 ] + num[ 1 ] * 10 + num[ 2 ] * 100 + num[ 3 ] * 1000 ;
                 
                  if ((next > 1000 && visited[next] != 1 ) && isprimer(next))
                 {
                     q.push(next);
                     result[next] = result[temp] + 1 ;
                     visited[next] = 1 ;                                             
                 }
                  if (next == end)
                 {
                    return  result[next];           
                 }
              }     
            } 
    }
     return  MAX;
}
int  main() {
     // freopen("s.txt","r",stdin);
     // freopen("key.txt","w",stdout);
     int  n,k,j,m;
    cin >> j;
     while (j -- )
    {
    memset(visited, 0 , sizeof (visited));
    memset(result, 0 , sizeof (result));          
    cin >> n >> k;
    m = BFS(n,k);
     if (m != MAX)          
    cout << m << endl;
     else
    cout << " Impossible " << endl;
    }
     return   0 ;
}

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

代码完全可以优化，比如定义一个结构体可以省掉result数组！
懒得改了
发现poj比joj更惊心动魄啊！经常compling,结果半天才出现！