Prime Path（POJ 3126 BFS）

Prime Path

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

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
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);
     }
 }