（广度搜索）A - Prime Path（11.1.1）

A - Prime Path（11.1.1）

Time Limit:1000MS    Memory Limit:65536KB    64bit IO Format:%I64d & %I64u

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Description

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 <iostream>
#include <cstring>
#include <queue>
using namespace std;
bool num[10001],used[10001];
int s,e;
void prim()//判断素数
{
	memset(num,true,sizeof(num));
	for(int i=2;i*i<10001;i++)
	{
		if(num[i])
		{
			for(int j=i+i;j<10000;j=j+i)
				num[j]=false;
		}
	}
	memset(num,false,1000);
}
struct node
{
	int x,step;
};
int d[4]={1,10,100,1000};//位数
int bfs(int x)
{
	queue<node> que;
	node tp,num;
	int tx;
	num.x=x,num.step=0;
	used[x]=false,que.push(num);
	while(!que.empty())
	{
		tp=que.front();
		que.pop();
		for(int i=0;i<4;i++)
			for(int j=0;j<=9;j++)
				if(i==3&&j==0)
					continue;
				else
				{
					if(j!=(tp.x/d[i]%10))
					{
						tx=((tp.x/d[i]/10)*10+j)*d[i]+tp.x%d[i];
						if(used[tx])
						{
							used[tx]=false,num.x=tx,num.step=tp.step+1,que.push(num);
							if(num.x==e)
								return num.step;
						}
					}
				}

	}
}
int main()
{
	int t;
	cin>>t;
	prim();
	while(t--)
	{
		cin>>s>>e;
		memcpy(used,num,10000);
		if(s==e)
			cout<<"0"<<endl;
		else cout<<bfs(s)<<endl;
	}
	return 0;
}