Prime Path(BFS)

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

这个题目是要从一个初始数字开始，每次改变一个数字，并且要求改变之后的这个数是素数，求每次这样进行重复的操作，直到达到目标数，求最少进行多少多少步，如果没法进行，那么就输出   Impossible   

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <queue>
using namespace std;
int t ,m ,n;
const int M = 10000;
int check[M];
int book[M];
int prime[M];
int ans;///标记是否找不到
struct stu
{
    int x;///存储每一个数
    int step;///记录走过的步数
}now,ne;
void make_prime()///此函数目的是找出哪些 是素数，哪些不是素数
{
    int tot=0;
    for(int i=2;i<=M;i++)
    {
        if(check[i]==0)
        {
            if(i>1000)
                prime[tot++]=i;
            for(int j=2*i;j<=M;j+=i)
                check[j]=1;
        }
    }
}

void bfs()
{
    queue<stu>Q;
    now.x=n;
    now.step=0;
    Q.push(now);
    while(!Q.empty())
    {
        now=Q.front();
        Q.pop();
        int d[4],i=0,t=now.x;
        while(t!=0)
        {
            d[i++]=t%10;///将原来的 数 各个位拆开，倒着存放在数组中。
            t/=10;
        }
        for(int i=0;i<4;i++)
        {
            for(int j=0;j<=9;j++)
            {
                if(j==d[i])continue;
                if(i==0&&(j%2==0||j==5))continue;///本身不是素数的，直接退出，末尾数字是偶数或者是5，都不是素数
                if(i==3&&j==0)continue;///第一位数字不是0
                int u=3,sum=0;
                while(u!=-1)///这一块太巧妙了，用了for循环将原来的数只改了一个数字成为了另一个数
                {
                    if(u==i)
                        sum=sum*10+j;
                    else sum=sum*10+d[u];
                    u--;
                }
                if(check[sum]==1||book[sum]==1)continue;
                book[sum]=1;
                ne.x=sum,ne.step=now.step+1;
                if(m==ne.x)
                {
                    ans=1;
                    printf("%d\n",ne.step);
                    return;
                }
                Q.push(ne);
            }
        }
    }
}
int main()
{
    make_prime();///因为每次判断改变之后的数是否是素数，所以用素数筛选法，提前先判断好哪些不是素数。
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        memset(book,0,sizeof(book));
        if(n==m)
            printf("0\n");
        else
        {
            ans=0;
            bfs();
            if(ans==0)
                printf("Impossible\n");
        }
    }
}