Poj 3126 -- prime path

Prime path

Time limit:1000 ms		Memory limit:65536 K
Total submissions:11288		Accepted:6398

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

Question: give two prime numbers and ask how many steps are required to convert the first one to the second one. Only one digit can be converted at a time. And the first digit cannot be zero.

Idea: A wide search can enumerate 9 or 10 cases for each digit. Question.

1/* ============================================== ====================================== 2 * Author: kevin 3 * filename: primepath. CPP 4 * creat time: * description: 6 ================================================ ===================================== */7 # include <iostream> 8 # include <algorithm> 9 # include <cstdio> 10 # include <cstring> 11 # include <queue> 12 # include <cmath> 13 # define CLR (, b) me Mset (a, B, sizeof (A) 14 # define M 15000 15 using namespace STD; 16 int isprime [M + 5], vis [m], CNT [m]; 17 int A, B; 18 void makeprime () 19 {20 CLR (isprime, 0); 21 isprime [0] = isprime [1] = 1; 22 For (INT I = 2; I <m; I ++) {23 if (! Isprime [I]) {24 for (Int J = I + I; j <m; j + = I) {25 isprime [J] = 1; 26} 27} 28} 29} 30 void BFS (INT s) 31 {32 CLR (VIS, 0); 33 CLR (CNT, 0 ); 34 vis [s] = 1; 35 queue <int> que; 36 que. push (s); 37 while (! Que. empty () {38 int T = que. front (); 39 que. pop (); 40 if (t = B) {41 printf ("% d \ n", CNT [T]); 42 break; 43} 44/* ------------------ handle 1st bits ---------------- */45 int No1 = T/1000; 46 int temp = T-No1 * 1000; 47 for (Int J = 1; j <= 9; j ++) {48 if (j = No1) continue; 49 int change_num = J * 1000 + temp; 50 if (! Isprime [change_num] &! Vis [change_num]) {51 que. push (change_num); 52 vis [change_num] = 1; 53 CNT [change_num] = CNT [T] + 1; 54} 55} 56/* --------------------- end ---------------------- */57/* -------------------- process 2nd-bit ----------------- */58 int NO2 = T/100% 10; 59 int S2 = T % 100; 60 for (Int J = 0; j <= 9; j ++) {61 If (j = NO2) continue; 62 int change_num = No1 * 1000 + J * 100 + S2; 63 If (! Isprime [change_num] &! Vis [change_num]) {64 que. push (change_num); 65 vis [change_num] = 1; 66 CNT [change_num] = CNT [T] + 1; 67} 68} 69/* --------------------- end ---------------------- */70/* -------------------- process 3rd bits ------------------- */71 int NO3 = T/10% 10; 72 S2 = T/100; 73 int S1 = T % 10; 74 for (Int J = 0; j <= 9; j ++) {75 if (j = NO3) continue; 76 int change_num = S2 * 100 + J * 10 + S1; 77 If (! Isprime [change_num] &! Vis [change_num]) {78 que. push (change_num); 79 vis [change_num] = 1; 80 CNT [change_num] = CNT [T] + 1; 81} 82} 83/* --------------------- end --------------------- */84/* ------------------- handle 4th-bit ------------------- */85 int No4 = T % 10; 86 int S3 = t-no4; 87 for (Int J = 0; j <= 9; j ++) {88 If (j = No4) continue; 89 int change_num = S3 + J; 90 if (! Isprime [change_num] &! Vis [change_num]) {91 que. push (change_num); 92 vis [change_num] = 1; 93 CNT [change_num] = CNT [T] + 1; 94} 95} 96/* -------------------- end ---------------------- */97} 98} 99 int main (INT argc, char * argv []) 100 {101 makeprime (); 102 int T; 103 scanf ("% d", & T); 104 while (t --) {105 scanf ("% d", & A, & B ); 106 BFS (a); 107} 108 return 0; 109}

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