題目大意是讓你求素數x到素數y的最短路徑距離，每次只能對素數中的一位數進行改變，每次改變1英鎊，求最小花費。 則素數a-b相連當且僅當a與b有一位數字不同。從起點開始bfs，若終點可達，即得到答案，不可達則輸出Impossible。 Prime Path

Time Limit: 1000MS	Memory Limit: 65536K
Total Submissions: 7281	Accepted: 4128

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

Source

#include <iostream>
#include <string.h>
#include <math.h>
#include <queue>
using namespace std;
int isprime( int n){
    int i,t;
    t=sqrt(n+0.0);
    for (i=2;i<=t;i++){
        if (n%i==0) return -2;
    }
    return -1;
}//prime:-1; not prime:-2; arrived:>=0;
const int S=10050;
int prime[S],f[S],w[4];
int main(){
    int i,T,x,y,tmp,a,b,c,d,t;
    for (i=2;i<S;i++) prime[i]=isprime(i);
    cin>>T;
    while (T--){
        cin>>x>>y;
        memcpy(f,prime,S*sizeof(int));
        f[x]=0;
        queue<int> q;
        q.push(x);
        while (!q.empty()&&(f[y]==-1))
        {
            t=q.front();
            tmp=t;
            q.pop();

            for (i=0;i<=3;i++) {w[i]=tmp%10;tmp/=10;}
            //ge_wei
            for (i=0;i<=9;i++)
            {
                tmp=i+10*w[1]+100*w[2]+1000*w[3];
                if (f[tmp]==-1) {
                    f[tmp]=f[t]+1;
                    q.push(tmp);
                }
            }
            //shi_wei
            for (i=0;i<=9;i++)
            {
                tmp=w[0]+10*i+100*w[2]+1000*w[3];
                if (f[tmp]==-1) {
                    f[tmp]=f[t]+1;
                    q.push(tmp);
                }
            }
            for (i=0;i<=9;i++)
            {
                tmp=w[0]+10*w[1]+100*i+1000*w[3];
                if (f[tmp]==-1) {
                    f[tmp]=f[t]+1;
                    q.push(tmp);
                }
            }
            for (i=1;i<=9;i++)
            {
                tmp=w[0]+10*w[1]+100*w[2]+1000*i;
                if (f[tmp]==-1) {
                    f[tmp]=f[t]+1;
                    q.push(tmp);
                }
            }

        }
        if (f[y]!=-1) cout<<f[y]<<endl;else cout<<"Impossible\n";
    }



    return 0;
}