Poj3126 prime path search BFS

Question:
Each time, only one digit can be changed to a new prime number,
How many times can be used to change to the second given prime number?
Analysis:
Since it is classified into BFs, we can use BFs to do it. We can judge a table with a prime number,
Speed up the determination, and then enumerate the number of each bit, the same as the previous BFs,
Calculate 0 to 9 each time. (Note: the value of a thousand bits cannot be 0. If the value of a single bits is an even number
You don't have to worry about it)

# Include <iostream>
# Include <cstring>
# Include <queue>
Using namespace STD;
Int m, n, ans;
Int A [6];
Bool isprime [10000], use [10000];
Void make_prime ()
{// Generate an array of prime numbers within 10000. If it is true, it is not a prime number.
Memset (isprime, false, sizeof (isprime ));
For (INT I = 3; I <= 9999; I + = 2)
If (! Isprime [I])
For (Int J = I + I; j <10000; j + = I)
Isprime [J] = true;
}
Void BFS ()
{
Int X;
Queue <int> P; // record the current number of times
Queue <int> q; // record the current four-digit number
P. Push (0 );
Q. Push (m );
While (! Q. Empty ())
{
Int T = Q. Front ();
Ans = P. Front ();
Q. Pop ();
P. Pop ();
If (t = N) // ends when the request is successful
Return;
For (INT I = 1; I <= 4; I ++) // records each of the four digits.
{
A [I] = T;
T/= 10;
}
For (Int J = 1; j <= 9; j ++)
{// The number of thousands is changing, and the number of thousands cannot be 0.
X = J * 1000 + A [3] * 100 + A [2] * 10 + A [1]; // current number
If (! Isprime [x] &! Use [x])
{// It is a prime number and is out of date.
Q. Push (X );
P. Push (ANS + 1 );
Use [x] = true; // indicates that you have tried it.
}
}
For (Int J = 0; j <= 9; j ++)
{// Hundreds of digits are changing
X = A [4] * 1000 + J * 100 + A [2] * 10 + A [1];
If (! Isprime [x] &! Use [x])
{
Q. Push (X );
P. Push (ANS + 1 );
Use [x] = true;
}
}
For (Int J = 0; j <= 9; j ++)
{// 10 is changing
X = A [4] * 1000 + A [3] * 100 + J * 10 + A [1];
If (! Isprime [x] &! Use [x])
{
Q. Push (X );
P. Push (ANS + 1 );
Use [x] = true;
}
}
For (Int J = 1; j <= 9; j + = 2)
{// The change of a single digit starts from 1 and increases by 2 at a time
X = A [4] * 1000 + A [3] * 100 + A [2] * 10 + J;
If (! Isprime [x] &! Use [x])
{
Q. Push (X );
P. Push (ANS + 1 );
Use [x] = true;
}
}
}
}
Int main ()
{
Freopen ("sum. In", "r", stdin );
Freopen ("sum. Out", "W", stdout );
Make_prime ();
Int T;
Cin> T;
While (t --)
{
Memset (use, false, sizeof (use ));
Ans = 0;
Cin> m> N;
BFS ();
Cout <ans <Endl;
}
Return 0;
}