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

题目大意：多组测试数据。给你两个素数n，m，要求每次只改变n的一个数字，且改变后的数仍然是素数，问通过多少次改变后素数n能变成素数m，输出变化最少的次数
思路：由于是求最小次数，用bfs能够保证。因此分别枚举个十百千位的数字+素数断定。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
using namespace std;
int t,n,m;
const int N=10010;
int vis[N];
struct node
{
    int x,step;
};
queue<node> q;
int pdss(int n)  //素数断定 
{
    if(n==0||n==1)
    return 0;
    else if(n==2||n==3)
    return 1;
    for(int i=2;i*i<=n;++i)
    {
        if(n%i==0)
        return 0;
    }
    return 1;
}
void bfs()
{
    int X,STEP,i,k;
    node ss,sss;
    while(!q.empty())
    {
        ss=q.front();
        q.pop();
        X=ss.x;
        STEP=ss.step;
        if(X==m)
        {
            printf("%d\n",STEP);
            return;
        }
        for(i=1;i<=9;i+=2) 
        {
            k=X/10*10+i;
            if(pdss(k)&&!vis[k]&&k!=X)
            {
                vis[k]=1;
                sss.x=k;
                sss.step=STEP+1;
                q.push(sss);            
            }
        }
        for(i=0;i<=9;i++) 
        {
            k=X/100*100+i*10+X%10;
            if(pdss(k)&&!vis[k]&&k!=X)
            {
                vis[k]=1;
                sss.x=k;
                sss.step=STEP+1;
                q.push(sss);            
            }
        }
        for(i=0;i<=9;i++) 
        {
            k=X/1000*1000+i*100+X%100;
            if(pdss(k)&&!vis[k]&&k!=X)
            {
                vis[k]=1;
                sss.x=k;
                sss.step=STEP+1;
                q.push(sss);            
            }
        }
        for(i=1;i<=9;i++) 
        {
            k=i*1000+X%1000;
            if(pdss(k)&&!vis[k]&&k!=X)
            {
                vis[k]=1;
                sss.x=k;
                sss.step=STEP+1;
                q.push(sss);            
            }
        }
    }
    printf("Impossible\n");
    return ;
}
int main(int argc, char *argv[])
{
    scanf("%d",&t);
    while(t--)
    {
        while(!q.empty()) q.pop();
        scanf("%d%d",&n,&m);
        memset(vis,0,sizeof(vis));
        node s;
        s.x=n;
        s.step=0;
        vis[n]=1;
        q.push(s);
        bfs();
    }
    return 0;
}