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

题意：给定两个素数n和m，要求把n变成m，每次变换时只能变一个数字，即变换后的数与变换前的数只有一个数字不同，并且要保证变换后的四位数也是素数。求最小的变换次数；如果不能完成变换，输出Impossible。

无论怎么变换，个位数字一定是奇数（个位数字为偶数肯定不是素数），这样枚举个位数字时只需枚举奇数就行；而且千位数字不能是0。所以可以用广搜，枚举各个数位上的数字，满足要求的数就加入队列，直到变换成功。因为是广搜，所以一定能保证次数最少。

注意：开始数字可以不是素数。
C++版本一

BFS

#include <iostream>
#include <stdio.h>
#include <queue>
#include <math.h>
#include <string.h>
int t,a,b;
int vis[10000];
using namespace std;
struct node{
    int x[4];
    int step;

}f,s,p;
bool prime(int n){
    for(int i=2;i<=(int)sqrt(n);i++){
        if(n%i==0)
            return false;
}

return true;
}
queue <node>Q;
int bfs(){
    while(!Q.empty()){
        Q.pop();
    }
    if(!prime(b)) return -1;

    if(a==b)return s.step;
    Q.push(s);
    while(!Q.empty()){
        f=Q.front();
        Q.pop();

        if(f.x[0]*1000+f.x[1]*100+f.x[2]*10+f.x[3]==b)
            return f.step;


        for(int i=0;i<4;i++){
            for(int j=0;j<=9;j++){
                if(f.x[i]!=j){
                    p=f;
                    p.x[i]=j;//printf("%d\n",p.x[0]*1000+p.x[1]*100+p.x[2]*10+p.x[3]);
                    int temp=p.x[0]*1000+p.x[1]*100+p.x[2]*10+p.x[3];
                    if(vis[temp]==0&&p.x[0]!=0&&prime(temp)){
                        p.step++;
                        vis[temp]=1;
                        Q.push(p);
                    }
                }
            }
        }
    }
return -1;
}
int main()
{
    scanf("%d",&t);
    while(t--){
        scanf("%d%d",&a,&b);
        memset(vis,0,sizeof(vis));
        s.x[0]=a/1000;
        s.x[1]=a%1000/100;
        s.x[2]=a%100/10;
        s.x[3]=a%10;
        s.step=0;
        vis[a]=1;
        int ans=bfs();

        if(ans>=0)printf("%d\n",ans);
        else printf("Impossible\n");
    }
    //cout << "Hello world!" << endl;
    return 0;
}

C++版本二

BFS

#include<cstdio>
#include<cstring>
#include<cmath>
#include<queue>
#include<algorithm>
using namespace std;
 
int n, m;
const int N = 1e4 + 100;
int vis[N];
struct node
{
    int x, step;
};
queue<node> Q;
 
bool judge_prime(int x) //判断素数
{
    if(x == 0 || x == 1)
        return false;
    else if(x == 2 || x == 3)
        return true;
    else
    {
        for(int i = 2; i <= (int)sqrt(x); i++)
            if(x % i == 0)
                return false;
        return true;
    }
}
 
void BFS()
{
    int X, STEP, i;
    while(!Q.empty())
    {
        node tmp;
        tmp = Q.front();
        Q.pop();
        X = tmp.x;
        STEP = tmp.step;
        if(X == m)
        {
            printf("%d\n",STEP);
            return ;
        }
        for(i = 1; i <= 9; i += 2) //个位
        {
            int s = X / 10 * 10 + i;
            if(s != X && !vis[s] && judge_prime(s))
            {
                vis[s] = 1;
                node temp;
                temp.x = s;
                temp.step = STEP + 1;
                Q.push(temp);
            }
        }
        for(i = 0; i <= 9; i++) //十位
        {
            int s = X / 100 * 100 + i * 10 + X % 10;
            if(s != X && !vis[s] && judge_prime(s))
            {
                vis[s] = 1;
                node temp;
                temp.x = s;
                temp.step = STEP + 1;
                Q.push(temp);
            }
        }
        for(i = 0; i <= 9; i++) //百位
        {
            int s = X / 1000 * 1000 + i * 100 + X % 100;
            if(s != X && !vis[s] && judge_prime(s))
            {
                vis[s] = 1;
                node temp;
                temp.x = s;
                temp.step = STEP + 1;
                Q.push(temp);
            }
        }
        for(i = 1; i <= 9; i++) //千位
        {
            int s = i * 1000 + X % 1000;
            if(s != X && !vis[s] && judge_prime(s))
            {
                vis[s] = 1;
                node temp;
                temp.x = s;
                temp.step = STEP + 1;
                Q.push(temp);
            }
        }
    }
    printf("Impossible\n");
    return ;
}
 
int main()
{
    int t, i;
    scanf("%d",&t);
    while(t--)
    {
        while(!Q.empty()) Q.pop();
        scanf("%d%d",&n,&m);
        memset(vis,0,sizeof(vis));
        vis[n] = 1;
        node tmp;
        tmp.x = n;
        tmp.step = 0;
        Q.push(tmp);
        BFS();
    }
    return 0;
}