poj3126（最爱的广搜）

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

题目意思是让你看第一个四位数变到第二个四位数要多少步，其中每变一步改动一个数字，改完数字之后的新的数还得是质数，求最小的改变次数，还有什么犹豫不决吗？果断广搜解决问题啊！！！

广搜重在理解过程，如果理解了会觉得非常简单！（虽然代码打起来复杂，我打个比方就是深搜是美女，却没内涵，广搜长得其丑无比，可是人家内涵比深搜丰富多了！！）

#include 
#include
#include
using namespace std;
typedef long long ll;
bool p[100000],vis[10000];
int a[4],b[4];
struct data
{
    int n,num;
}x[100000];
void get()
{
    memset(p,1,sizeof(p));
    for(ll i=2;i<100000;i++)
    {
        if(p[i]==1)
        for(ll j=i*i;j<100000;j+=i)
            p[j]=0;
    }
}

int hh(int *a)
{
    return a[0]*1000+a[1]*100+a[2]*10+a[3];
}

int bfs(int *a,int *b)
{
    if(hh(a)==hh(b)||p[hh(b)]==0)return 0;
    int l=0,r=0;
    vis[hh(a)]=1;
    x[r].n=hh(a);
    x[r++].num=0;
    while(l>c;
   while(c--)
   {
       memset(vis,0,sizeof(vis));
       int a0,b0;
       cin>>a0>>b0;
       a[0]=a0/1000,a[1]=(a0/100)%10,a[2]=(a0/10)%10,a[3]=a0%10;
       b[0]=b0/1000,b[1]=(b0/100)%10,b[2]=(b0/10)%10,b[3]=b0%10;
       int ss=bfs(a,b);
       if(ss==-1)cout<<"Impossible"<