D - Prime Path

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

代码：

#include<cstdio>
#include<iostream>
#include<string>
#include<queue>
#include<vector>
#include<cstring>
using namespace std;
vector<int>pr(10010,1);
vector<int>in(10010);
struct prime
{
	int x;
	int times;
	prime(int xx,int tt):x(xx),times(tt){
	}
};

int main(void)
{
	int t,x,y;
	scanf("%d",&t);
	int i,j;
	for(i=2;i<101;i++)
	{
		if(pr[i])
		{
			for(j=2*i;j<=10000;j+=i)
			{
				pr[j]=0;
			}
		}
	}
	while(t--)
	{
		scanf("%d %d",&x,&y);
		if(x==y){
		printf("0\n");
		continue;}
		in.assign(in.size(),0);//vector清0 
		queue<prime> q;
		q.push(prime(x,0));
		in[x]=1;
		while(q.size())
		{
			prime now=q.front();
			if(now.x==y)
			{
				break;
			}
			q.pop();
			int a=now.x%10;
			int b=(now.x%100)/10;
			int c=(now.x%1000)/100;
			int d=now.x/1000;
			int sum;
			for(i=0;i<=9;i++)
			{
				sum=d*1000+c*100+b*10+i;
				if(pr[sum]&&!in[sum])
				{
					q.push(prime(sum,now.times+1));
					in[sum]=1;
				}
			}
			for(i=0;i<=9;i++)
			{
				sum=d*1000+c*100+i*10+a;
				if(pr[sum]&&!in[sum])
				{
					q.push(prime(sum,now.times+1));
					in[sum]=1;
				}
			}
			for(i=0;i<=9;i++)
			{
				sum=d*1000+i*100+b*10+a;
				if(pr[sum]&&!in[sum])
				{
					q.push(prime(sum,now.times+1));
					in[sum]=1;
				}
			}
			for(i=1;i<=9;i++)//从1开始哦 
			{
				sum=i*1000+c*100+b*10+a;
				if(pr[sum]&&!in[sum])
				{
					q.push(prime(sum,now.times+1));
					in[sum]=1;
				}
			}
		}
		if(q.size())
		printf("%d\n",q.front().times); 
		else
		printf("Impossible\n");
	}
	return 0;
 }