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

31033 81791373 80171033 1033

Sample Output

670

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");        }    }}