Prime Path

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 15325		Accepted: 8634

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.

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).

One line for each case, either with a number stating the minimal cost or containing the word Impossible.

3

1033 8179

1373 8017

1033 1033

6

7

0
单纯的将所有情况都搜一遍

 #include <iostream>

 #include <cstring>

 #include <algorithm>

 #include <queue>

 using namespace std;

 bool prime[];

 bool vis[];

 int n[];

 int a,b;

 struct node

 {

     int num,t;

 };

 void init()

 {

     int i,j;

     memset(prime,,sizeof(prime));

     for(i=;i<;i++)

     {

         for(j=;j<i;j++)

         {

             if(i%j==)

                 break;

         }

         if(j==i)

         {

             prime[i]=;    //1代表质数

             //cout<<i<<endl;

         }

     }

     for(i=;i<;i++)

         n[i]=i;

     n[]=,n[]=,n[]=,n[]=;

     return;

 }

 int bfs()

 {

     int i,j;

     queue<node> Q;

     node tem,u;

     int k;

     tem.num=a,tem.t=;

     memset(vis,,sizeof(vis));

     Q.push(tem);

     while(!Q.empty())

     {

         tem=Q.front();

         Q.pop();

         //cout<<tem.num<<endl;

         /*if(tem.num==3733)

             printf("adsfadsf\n");*/

         if(tem.num==b)

             return tem.t;

         for(i=;i<=;i++)

         {

             if(tem.num%==n[i])

                 continue;

             u.num=tem.num/;

             u.num=u.num*+n[i];

             //cout<<u.num<<" "<<prime[u.num]<<" "<<vis[u.num]<<endl;

             if(vis[u.num]==&&prime[u.num])

             {

                 u.t=tem.t+;

                 vis[u.num]=;

                 Q.push(u);

             }

         }

         for(i=;i<=;i++)

         {

             k=(tem.num/)%;

             if(k==n[i])

                 continue;

             k=tem.num%;

             u.num=((tem.num/)*+n[i])*+k;

             if(vis[u.num]==&&prime[u.num])

             {

                 u.t=tem.t+;

                 vis[u.num]=;

                 Q.push(u);

             }

         }

         for(i=;i<=;i++)

         {

             k=(tem.num/)%;

             if(k==n[i])

                 continue;

             k=tem.num%;

             u.num=((tem.num/)*+n[i])*+k;

             if(vis[u.num]==&&prime[u.num])

             {

                 u.t=tem.t+;

                 vis[u.num]=;

                 Q.push(u);

             }

         }

         for(i=;i<=;i++)

         {

             k=tem.num/;

             if(k==n[i])

                 continue;

             //cout<<k<<" "<<n[i]<<endl;

             k=tem.num%;

             u.num=n[i]*+k;

         //    cout<<u.num<<endl;

             if(vis[u.num]==&&prime[u.num])

             {

                 u.t=tem.t+;

                 vis[u.num]=;

                 //cout<<u.num<<endl;

                 Q.push(u);

             }

         }

         //break;

     }

     return -;

 }

 int main()

 {

     int T,ans;

     init();

     freopen("in.txt","r",stdin);

     //freopen("ou.txt","w",stdin);

     scanf("%d",&T);

     while(T--)

     {

         scanf("%d%d",&a,&b);

         ans=bfs();

         if(ans==-)    printf("Impossible\n");

         else printf("%d\n",ans);

     }

 }