Prime Path

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

Sample Output

Source

Northwestern Europe 2006

 //2017-02-23

 #include <iostream>

 #include <cstdio>

 #include <cstring>

 #include <queue>

 using namespace std;

 struct node

 {

     int num, step;

 };

 bool isPrime(int n)

 {

     for(int i = ; i*i <= n; i++)

           if(n%i==)return false;

     return true;

 }

 int pow(int a, int n)

 {

     int ans = ;

     for(int i = ; i < n; i++)

         ans *= a;

     return ans;

 }

 int main()

 {

     int n, x, y;

     bool book[];

     cin>>n;

     while(n--)

     {

         cin>>x>>y;

         queue<node> q;

         node tmp;

         tmp.num = x;

         tmp.step = ;

         q.push(tmp);

         int num, step;

         bool ok = false;

         memset(book, , sizeof(book));

         book[x] = ;

         while(!q.empty()){

             num = q.front().num;

             step = q.front().step;

             q.pop();

             if(num == y){

                 ok = true;

                 cout<<step<<endl;

                 break;

             }

             for(int i = ; i < ; i++){

                 for(int p = ; p < ; p++){

                     if(i== && p==)continue;

                     int nowNum = num-((num/pow(, i))%)*pow(, i)+p*pow(, i);

                     if(!book[nowNum] && isPrime(nowNum)){

                         tmp.num = nowNum;

                         tmp.step = step+;

                         q.push(tmp);

                         book[nowNum] = ;

                     }

                 }

             }

             if(ok)break;

         }

     }

     return ;

 }

巴特西

POJ3126(KB1-F BFS)