GZHU18级寒假训练：Gemini's Trial H

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

#include<queue>
#include<cmath>
#include <iostream>
using namespace std;
struct Q
{
	int a[4], step;
	
}one ,two;
Q set(int n,Q ab)
{
	ab.a[0] = n/1000;
	n -=ab.a[0]*1000;
	ab.a[1] = n / 100;
	n -=ab.a[1]* 100;
	ab.a[2] = n / 10;
	ab.a[3] = n - ab.a[2]*10;
	ab.step = 0;
	return ab;
}
bool count(Q ab)
{
	int digital;
	digital = ab.a[0] * 1000 + ab.a[1] * 100 + ab.a[2] * 10 + ab.a[3];
	for (int i = 2; i*i <= digital; i++)
	{
		if (digital%i == 0)
		{
			return false;
		}
	}
	return true;
}
int vis[10][10][10][10];
void bfs()
{
	
	queue<Q> one2;
	one2.push(one);
	while (!one2.empty()) 
	{
		Q s;
		 s= one2.front();
		one2.pop();
		int a, b, c, d;
		a=s.a[0];
		b=s.a[1];
		c=s.a[2];
		d=s.a[3];
		vis[a][b][c][d] = 1;
		if (a == two.a[0] && b == two.a[1] && c == two.a[2] && d == two.a[3])
		{
			cout << s.step << endl;
			return;
		}
		for (int i = 1; i < 10; i ++)
		{
			Q e = s;
			e.a[0] = i;
			e.step++;
			if (!count(e)) continue;
			if (vis[i][b][c][d]) continue;
			vis[i][b][c][d] = 1;
			one2.push(e);
		}
		for (int i = 0; i < 10; i ++)
		{
			Q e = s;
			e.a[1] = i;
			e.step++;
			if (!count(e)) continue;
			if (vis[a][i][c][d]) continue;
			vis[a][i][c][d] = 1;
			one2.push(e);
		}
		for (int i = 0; i < 10; i++)
		{
			Q e = s;
			e.a[2] = i;
			e.step++;
			if (!count(e)) continue;
			if (vis[a][b][i][d]) continue;
			vis[a][b][i][d] = 1;
			one2.push(e);
		}
		for (int i = 1; i < 10; i+=2)
		{
			Q e = s;
			e.a[3] = i;
			e.step++;
			if (!count(e)) continue;
			if (vis[a][b][c][i]) continue;
			vis[a][b][c][i] = 1;
			one2.push(e);
		}
	}

	cout<<"Impossible"<<endl;
	return;
}
int main()
{
	int n;
	scanf_s("%d", &n);
	while (n--)
	{
		memset(vis, 0, sizeof(vis));
		int a,b;
		scanf_s("%d %d", &a,&b);
		one=set(a, one);
		two=set(b, two);
		bfs();
	}
   
}