Kattis之旅——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.

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 1	Sample Output 1
3 1033 8179 1373 8017 1033 1033	6 7 0

大致意思就是由前面的那个素数变到后面的那个素数，每次只能变一位数，变化后的数也应该是一个素数（无论是不是所求的数），求变化次数。

直接BFS即可。

 //Asimple

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

const int maxn = ;

ll n, m, s, res, ans, len, T, k;

int x, y;

int pr[maxn];

int P(int n) {

    for(int i=; i*i<=n; i++) {

        if( n%i== ) return ;

    }

    return ;

}

//将k位化0

int change(int n, int k) {

    char s[] = {};

    sprintf(s, "%d", n);

    s[k] = '';

    sscanf(s, "%d", &n);

    return n;

}

int solve(int s, int e) {

    queue<int> q;

    int dis[maxn] = {};

    q.push(s);

    dis[s] = ;

    while( q.size() ) {

        s = q.front();

        q.pop();

        if( s == e ) return dis[s]-;

        int t = ;

        for(int i=; i<; i++) {

            int k = change(s, i);

            for(int j=; j<; j++) {

                int a = k+j*t;

                if( pr[a]== && dis[a]== ) {

                    q.push(a);

                    dis[a] = dis[s]+;

                }

            }

            t /= ;

        }

    }

    return -;

}

void input() {

    for(int i=; i<maxn; i++) pr[i] = P(i);

    cin >> T;

    while( T -- ) {

        cin >> x >> y;

        ans = solve(x, y);

        if( ans==- ) puts("Impossible");

        else cout << ans << endl;

    }

}

int main(){

    input();

    return ;

}