POJ 3126 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

31033 81791373 80171033 1033

Sample Output

670

#include<iostream>#include<cstdio>#include<cstring>#include<queue>#include<cmath>using namespace std;int m,n;struct node{    int x,s;};queue<struct node>q;int vis[9999];int Judge(int n){    int i,j;    if(n==0||n==1) return 0;    else if(n==2||n==3) return 1;    else    {        for(i=2;i<=(int)sqrt(n);i++)        {            if(n%i==0) return 0;        }        return 1;    }}void BFS(){    int i;    while(!q.empty())    {        struct node t;        t=q.front();        q.pop();        if(t.x==m)        {            printf("%d\n",t.s);            return ;        }        for(i=1;i<=9;i+=2)        {            int x=t.x/10*10+i;            if(!vis[x]&&Judge(x)&&x!=t.x)            {                vis[x]=1;                struct node f;                f.x=x;                f.s=t.s+1;                q.push(f);            }        }        for(i=0;i<=9;i++)        {            int x=t.x/100*100+i*10+t.x%10;            if(!vis[x]&&Judge(x)&&x!=t.x)            {                vis[x]=1;                struct node f;                f.x=x;                f.s=t.s+1;                q.push(f);            }        }        for(i=0;i<=9;i++)        {            int x=t.x/1000*1000+i*100+t.x%100;            if(x!=t.x&&Judge(x)&&!vis[x])            {                vis[x]=1;                struct node f;                f.x=x;                f.s=t.s+1;                q.push(f);            }        }        for(i=1;i<=9;i++)        {            int x=i*1000+t.x%1000;            if(x!=t.x&&Judge(x)&&!vis[x])            {                vis[x]=1;                struct node f;                f.x=x;                f.s=t.s+1;                q.push(f);            }        }    }    printf("Impossible\n");    return ;}int main(){    int t;    scanf("%d",&t);    while(t--)    {        memset(vis,0,sizeof(vis));        while(!q.empty()) q.pop();        scanf("%d%d",&n,&m);        if(n==m) printf("0\n");        else        {            vis[n]=1;            struct node t;            t.x=n;            t.s=0;            q.push(t);            BFS();        }    }    return 0;}