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.

    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 “<tt class="ttfamily">Impossible</tt>”.

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 = 10000;
ll n, m, s, res, ans, len, T, k;
int x, y;
int pr[maxn];

int P(int n) {
    for(int i=2; i*i<=n; i++) {
        if( n%i==0 ) return 0;
    }
    return 1;
}
//将k位化0 
int change(int n, int k) {
    char s[6] = {0};
    sprintf(s, "%d", n);
    s[k] = '0';
    sscanf(s, "%d", &n);
    return n;
}

int solve(int s, int e) {
    queue<int> q;
    int dis[maxn] = {0};
    q.push(s);
    dis[s] = 1;
    while( q.size() ) {
        s = q.front();
        q.pop();
        if( s == e ) return dis[s]-1;
        int t = 1000;
        for(int i=0; i<4; i++) {
            int k = change(s, i);
            for(int j=0; j<10; j++) {
                int a = k+j*t;
                if( pr[a]==1 && dis[a]==0 ) {
                    q.push(a);
                    dis[a] = dis[s]+1;
                }
            }
            t /= 10;
        }
    }
    return -1;
}

void input() {
    for(int i=1000; i<maxn; i++) pr[i] = P(i);
    cin >> T;
    while( T -- ) {
        cin >> x >> y;
        ans = solve(x, y);
        if( ans==-1 ) puts("Impossible");
        else cout << ans << endl;
    }
}

int main(){
    input();
    return 0;
}