POJ3126(KB1-F BFS)

Prime Path

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

Sample Output

6
7
0

Source

Northwestern Europe 2006

 1 //2017-02-23
 2 #include <iostream>
 3 #include <cstdio>
 4 #include <cstring>
 5 #include <queue>
 6 
 7 using namespace std;
 8 
 9 struct node
10 {
11     int num, step;
12 };
13 
14 bool isPrime(int n)
15 {
16     for(int i = 2; i*i <= n; i++)
17           if(n%i==0)return false;
18     return true;
19 }
20 
21 int pow(int a, int n)
22 {
23     int ans = 1;
24     for(int i = 0; i < n; i++)
25         ans *= a;
26     return ans;
27 }
28 
29 int main()
30 {
31     int n, x, y;
32     bool book[10005];
33     cin>>n;
34     while(n--)
35     {
36         cin>>x>>y;
37         queue<node> q;
38         node tmp;
39         tmp.num = x;
40         tmp.step = 0;
41         q.push(tmp);
42         int num, step;
43         bool ok = false;
44         memset(book, 0, sizeof(book));
45         book[x] = 1;
46         while(!q.empty()){
47             num = q.front().num;
48             step = q.front().step;
49             q.pop();
50             if(num == y){
51                 ok = true;
52                 cout<<step<<endl;
53                 break;
54             }
55             for(int i = 0; i < 4; i++){
56                 for(int p = 0; p < 10; p++){
57                     if(i==3 && p==0)continue;
58                     int nowNum = num-((num/pow(10, i))%10)*pow(10, i)+p*pow(10, i);
59                     if(!book[nowNum] && isPrime(nowNum)){
60                         tmp.num = nowNum;
61                         tmp.step = step+1;
62                         q.push(tmp);
63                         book[nowNum] = 1;
64                     }
65                 }
66             }
67             if(ok)break;
68         }
69     }
70 
71     return 0;
72 }