POJ 3126 Prime Path

Prime Path

https://vjudge.net/problem/POJ-3126
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

C++

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>

using namespace std;

const int MAXN = 10000 + 10;
bool isPrime[MAXN];
int prime[MAXN];
bool vis[MAXN];

void euler()
{
	int cnt = 0;
	for(int i = 2; i < MAXN; i ++){
		if(!isPrime[i])
			prime[cnt ++] = i;
		for(int j = 0; j < cnt && i * prime[j] < MAXN; j ++)
		{
			isPrime[i * prime[j]] = true;
			if(i % prime[j] == 0) break;
		}
	}
}

struct node{
	int num, step;
};

int bfs(int n, int m)
{
	queue<node> q;
	node now, next;
	now.num = n;
	now.step = 0;
	q.push(now);
	vis[n] = true;
	int a, b, c, d;
	while(!q.empty())
	{
		now = q.front();
		q.pop();
		if(now.num == m)
			return now.step;
		d = now.num % 10;
		c = now.num / 10 % 10;
		b = now.num / 100 % 10;
		a = now.num / 1000;
		next.step = now.step + 1;
		for(int i = 0; i < 4; i ++)
		{
			for(int j = 0; j <= 9; j ++)
			{
				if(i == 0 && j == 0) continue;
				if(i == 0)
					next.num = j * 1000 + b * 100 + c * 10 + d;
				else if(i == 1)
					next.num = a * 1000 + j * 100 + c * 10 + d;
				else if(i == 2)
					next.num = a * 1000 + b * 100 + j * 10 + d;
				else if(i == 3)
					next.num = a * 1000 + b * 100 + c * 10 + j;
				if(!isPrime[next.num] && !vis[next.num])
				{
					vis[next.num] = true;
					q.push(next);
				}
			}
		}		
	}
	return -1;
}

int main()
{
	int t, n, m;
	scanf("%d", &t);
	euler();
	while(t --)
	{
		scanf("%d%d", &n, &m);
		memset(vis, false, sizeof(vis));
		int ans = bfs(n, m);
		if(ans == -1) printf("Impossible\n");
		else printf("%d\n", ans);
	}
	return 0;
}