F - Prime Path

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

#include<iostream>
#include<queue>
#include<cstring>
using namespace std;
typedef struct
{
	int a[4],step;
}N;
bool flag;
int vis[10000];
bool judge(int n)
{
	for(int i=2;i*i<=n;i++)
        if(n%i==0)
			return false;
		return true;
}
void bfs(N s,int m)
{
	queue<N>q;
	q.push(s);
	while(q.size())
	{
		N tm=q.front(),tm1;
		q.pop();
		int mm=tm.a[0]*1000+tm.a[1]*100+tm.a[2]*10+tm.a[3];
		if(mm==m)
		{
			cout<<tm.step<<endl;
			flag=true;
			break;
		}
		tm1=tm;
		for(int k=1;k<=9;k++)
		{
          tm1.a[0]=k;
		  int t=tm1.a[0]*1000+tm1.a[1]*100+tm1.a[2]*10+tm1.a[3];
		  if(judge(t)&&!vis[t])
		  {
			  vis[t]=1;
			  tm1.step=tm.step+1;
			  q.push(tm1);
		  }
		  tm1=tm;
		}
		for(int i=0;i<=9;i++)
		{
			for(int j=1;j<4;j++)
			{
				tm1.a[j]=i;
				int t=tm1.a[0]*1000+tm1.a[1]*100+tm1.a[2]*10+tm1.a[3];
              if(judge(t)&&!vis[t])
			  {
			  vis[t]=1;
			  tm1.step=tm.step+1;
			  q.push(tm1);
			  }
		     tm1=tm;
			}
		}
	}
}
int main()
{
	int n,m,t;
	cin>>t;
	while(t--)
	{
		cin>>n>>m;
		memset(vis,0,sizeof(vis));
		flag=false;
		N s;
		s.a[0]=n/1000;
		s.a[1]=(n/100)%10;
		s.a[2]=(n/10)%10;
		s.a[3]=n%10;
		s.step=0;
		bfs(s,m);
		if(!flag)
		cout<<"Impossible"<<endl;
	}
	return 0;
}