Prime Path(素数筛选+bfs)

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9519		Accepted: 5458

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
题意：给定两个四位数，求从前一个数变到后一个数最少需要几步，改变的原则是每次只能改变某一位上的一个数，而且每次改变得到的必须是一个素数；

思路：将四位数以内的素数筛选出来，bfs时，枚举四位数的每一位上的每一个数；

 #include<stdio.h>
 #include<string.h>
 #include<queue>
 using namespace std;

 struct node
 {
     int num;
     int step;
 };
 queue<struct node>que;
 int n,m,flag;
 bool p[],vis[];

 //素数筛；
 void prime_search()
 {
     memset(p,,sizeof(p));
     for(int i = ; i <= ; i+=)
         p[i] = ;
     for(int i = ; i <= ; i++)
     {
         if(p[i])
         {
             for(int j = i+i; j <= ; j += i)
                 p[j] = ;
         }
     }
 }

 int bfs()
 {
     while(!que.empty())
         que.pop();
     que.push((struct node){n,});
     vis[n] = ;
     int res,tmp,r,t,s;
     while(!que.empty())
     {
         struct node u = que.front();
         que.pop();
         if(u.num == m)
             return u.step;

         //枚举个位数
         r = u.num%;
         for(int k = -; k <= ; k++)
         {
             t = r+k;
             if(t >=  && t <= )
             {
                 res = (u.num/)*+t;
                 if(p[res] && !vis[res])
                 {
                     que.push((struct node){res,u.step+});
                     vis[res] = ;
                 }
             }
         }

         //枚举十位数
         tmp = u.num/;
         r = tmp%;
         s = tmp/;
         for(int k = -; k <= ; k++)
         {
             t = r+k;
             if(t >=  && t <= )
             {
                 res = (s*+t)*+u.num%;
                 if(p[res] && !vis[res])
                 {
                     que.push((struct node){res,u.step+});
                     vis[res] = ;
                 }
             }
         }

         //枚举百位数
         int tmp = u.num/;
         r = tmp%;
         s = tmp/;
         for(int k = -; k <= ; k++)
         {
             t = r+k;
             if(t >=  && t <= )
             {
                 res = (s*+t)*+u.num%;
                 if(p[res] && !vis[res])
                 {
                     que.push((struct node){res,u.step+});
                     vis[res] = ;
                 }
             }
         }

         //枚举千位数
         r = u.num/;
         for(int k = -; k <= ; k++)
         {
             t = r+k;
             if(t > && t <= )
             {
                 res = t*+u.num%;
                 if(p[res] && !vis[res])
                 {
                     que.push((struct node){res,u.step+});
                     vis[res] = ;
                 }
             }
         }
     }
 }
 int main()
 {
     int test;
     prime_search();
     scanf("%d",&test);
     while(test--)
     {
         memset(vis,,sizeof(vis));
         scanf("%d %d",&n,&m);
         int ans = bfs();
         printf("%d\n",ans);
     }
     return ;
 }