Prime Path（poj 3126)

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

题目大意：问输入的第一个数金过几次变换可以得到第二个数；
变换时，每次只能改变一个数字；
经过变换得到的数字必须是素数；
不能完成输出Impossible;

#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
using namespace std;
bool pr[10003],vis[10003];
int a,b,t,i,j;
struct node
{
    int a,step;
}p,q;
void pri()
{
    memset(pr,-1,sizeof(pr));
    pr[0]=pr[1]=0;
    for(i=2; i<10003; i++)
    {
        if(pr[i])
            for(j=2*i; j<10003; j+=i)
                pr[j]=0;
    }
}
int change(int x,int i,int j)
{//方便改变数字的每一位，x是原数字，i代表第几位i=1是个位，j是改编成几（0————9，千位不能为0）
    if(i==1) return (x/10)*10+j;
    else if(i==2) return (x/100)*100+x%10+j*10;
    else if(i==3) return (x/1000)*1000+x%100+j*100;
    else if(i==4) return (x%1000)+j*1000;
}
void bfs()//简单bfs
{
    queue<node>que;
    p.a=a;
    p.step=0;
    vis[a]=1;
    que.push(p);
    while(!que.empty())
    {
        p=que.front();
        que.pop();
        q.step=p.step+1;
        for(i=1; i<5; i++)
            for(j=0; j<10; j++)
            {
                if(i==4&&j==0)
                    continue;
                q.a=change(p.a,i,j);
                if(q.a==b)
                {
                    printf("%d\n",q.step);
                    return;
                }
                if(pr[q.a]&&!vis[q.a])
                {
                    que.push(q);
                    vis[q.a]=1;
                }
            }
    }
    printf("Impossible\n");
}
int main()
{
    pri();//素数筛初始化
    scanf("%d",&t);
    while(t--)
    {
        memset(vis,0,sizeof(vis));//初始化
        scanf("%d %d",&a,&b);
        if(a==b){printf("0\n");continue;}//a==b情况单独处理；
        bfs();
    }
    return 0;
}