POJ 3126 Prime Path (Guang suo)

POJ 3126 Prime Path (Guang suo)
Prime Path

Time Limit:1000 MS		Memory Limit:65536 K
Total Submissions:12974		Accepted:7342

Description

The ministers of the cabinet were quite upset by the message from the Chief of Security stating that they wowould 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! Also 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

31033 81791373 80171033 1033

Sample Output

670

Give two prime numbers a and B (4 digits) and ask if a can be transformed to B. Conversion principle: Only one of the digits of a can be changed at a time, and the converted number must also be a prime number. For example, 1033 can be converted to 1733, but not 1233 (because 1233 is not a prime number ), it cannot be changed to 3733 (because 2 digits are changed from 1033 to 3733 at a time). If yes, the minimum number of transformations is output; otherwise, Impossible is output.
You can perform extensive search to determine whether a table is a prime number.

# Include
 
  
# Include
  
   
# Include
   
    
Using namespace std; # define HUR 100 # define THO 1000 # define TEN 10 const int maxn = 10000; bool prime [maxn + 5]; int vis [maxn], s, e; void prime_table () {int I, j; memset (prime, 0, sizeof (prime); for (I = 2; I
    
     
Que; que. push (s); while (! Que. empty () {int t = que. front (); que. pop (); int d = t; d % = 1000; for (I = 1; I <10; I ++) {int tt = d + I * THO; // transformer if (prime [tt] = 0 & vis [tt] = 0) {if (tt = e) return vis [t]; que. push (tt); vis [tt] = vis [t] + 1 ;}} d = t % 100 + (t/1000*1000); for (I = 0; I <10; I ++) {int tt = d + I * HUR; // transform the hundred-bit if (prime [tt] = 0 & vis [tt] = 0) {if (tt = e) return vis [t]; que. push (tt); vis [tt] = vis [t] + 1 ;}} d = t % 10 + t/100*100; for (I = 0; I <10; I ++) {int tt = d + I * TEN; // convert the ten-digit if (prime [tt] = 0 & vis [tt] = 0) {if (tt = e) return vis [t]; que. push (tt); vis [tt] = vis [t] + 1 ;}} d = t/10*10; for (I = 0; I <10; I ++) {int tt = d + I; // transform a single bit if (prime [tt] = 0 & vis [tt] = 0) {if (tt = e) return vis [t]; que. push (tt); vis [tt] = vis [t] + 1 ;}}return 0 ;}int main () {int T, res; prime_table (); scanf ("% d", & T); while (T --) {scanf ("% d", & s, & e); if (s = e) {printf ("0 \ n"); continue;} res = bfs (); if (res = 0) printf ("Impossible \ n "); elseprintf ("% d \ n", res);} return 0 ;}