Prime Path（poj 3126)

Description

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：问输入的第一个数金过几次变换可以得到第二个数；
变换时，每次只能改变一个数字；
经过变换得到的数字必须是素数；
不能完成输出Impossible;

 #include<iostream>
 #include<cstring>
 #include<cstdio>
 #include<queue>
 using namespace std;
 bool pr[],vis[];
 int a,b,t,i,j;
 struct node
 {
     int a,step;
 }p,q;
 void pri()
 {
     memset(pr,-,sizeof(pr));
     pr[]=pr[]=;
     for(i=; i<; i++)
     {
         if(pr[i])
             for(j=*i; j<; j+=i)
                 pr[j]=;
     }
 }
 int change(int x,int i,int j)
 {//方便改变数字的每一位，x是原数字，i代表第几位i=1是个位，j是改编成几（0————9，千位不能为0）
     if(i==) return (x/)*+j;
     else if(i==) return (x/)*+x%+j*;
     else if(i==) return (x/)*+x%+j*;
     else if(i==) return (x%)+j*;
 }
 void bfs()//简单bfs
 {
     queue<node>que;
     p.a=a;
     p.step=;
     vis[a]=;
     que.push(p);
     while(!que.empty())
     {
         p=que.front();
         que.pop();
         q.step=p.step+;
         for(i=; i<; i++)
             for(j=; j<; j++)
             {
                 if(i==&&j==)
                     continue;
                 q.a=change(p.a,i,j);
                 if(q.a==b)
                 {
                     printf("%d\n",q.step);
                     return;
                 }
                 if(pr[q.a]&&!vis[q.a])
                 {
                     que.push(q);
                     vis[q.a]=;
                 }
             }
     }
     printf("Impossible\n");
 }
 int main()
 {
     pri();//素数筛初始化
     scanf("%d",&t);
     while(t--)
     {
         memset(vis,,sizeof(vis));//初始化
         scanf("%d %d",&a,&b);
         if(a==b){printf("0\n");continue;}//a==b情况单独处理；
         bfs();
     }
     return ;
 }