題目連結：https://vjudge.net/problem/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

3
1033 8179
1373 8017
1033 1033

Sample Output

6
7
0

題目大意：給出兩個四位的素數a，b。
a可以通過改變某一位上的數字使其變成c，但只有當c也是四位的素數時才能進行這種改變。對於每個樣例，輸出最少變換次數，如果無法變換成b則輸出"Impossible"。

解題思路：利用BFS對素數的各位進行替換，替換後的數為素數時才壓入佇列，這樣找出的次數即為最少次數。

#include<iostream>
#include<queue>
#include<cmath>
#include<cstring>
using namespace std;
bool vis[10010], is_primer[10010];//標記陣列
int step[10010];//轉換步數

bool Is_pri(int x){//判斷素數
    for (int i = 2; i <= sqrt(x*1.0);i++){
        if(x%i==0){
            return 0;
        }
    }
    return 1;
}

int bfs(int xx,int y){
    memset(vis, 0, sizeof(vis));//初始化
    memset(step, 0, sizeof(step));
    int a[4];
    queue<int> q;
    vis[xx] = 1;//訪問後標記
    q.push(xx);
    while(!q.empty()){
        int x = q.front();
        q.pop();
        if(x==y){
            return step[x];
        }
        a[0] = x / 1000;
        a[1] = (x / 100) % 10;
        a[2] = (x / 10) % 10;
        a[3] = x % 10;
    for (int j = 0; j < 4;j++){//一位一位的替換
         int num = a[j];
        for (int i = 0; i < 10;i++){
            if(i!=num){//選取不是原來的數進行替換
                a[j] = i;
                int sum = a[0] * 1000 + a[1] * 100 + a[2] * 10 + a[3];
                if(!vis[sum]&&is_primer[sum]){
                    vis[sum] = 1;
                    q.push(sum);
                    step[sum] = step[x] + 1;
                }
            }
        }
        a[j] = num;
    }
    }
    return -1;
}
int main(){
    for (int i = 1000; i < 10000;i++){
        is_primer[i] = Is_pri(i);
    }
    int n;
    cin >> n;
    while (n--)
    {
        int a, b;
        cin >> a >> b;
        if(bfs(a,b)==-1){
            cout << "Impossible" << endl;
        }
        else{
            cout << bfs(a, b) << endl;
        }
        /* code */
    }
    return 0;
}

Java Code

import java.util.LinkedList;
import java.util.Queue;
import java.util.Scanner;



public class Main {
	public static int maxn=(int)1e5+10;
	static boolean vis[]=new boolean[maxn];//標記陣列
	static boolean isprim[]=new boolean[maxn];//素數標記陣列
	static void Era(int n) {//Eratosthenes篩法
		for(int i=2;i<=n;i++) {
			if(!vis[i]) {
				isprim[i]=true;
				vis[i]=true;
				for(int j=(i+i);j<=n;j+=i) {
					vis[j]=true;
				}
			}
			for(int k=0;k<1000;k++) {//只能轉換成4位數
				isprim[k]=false;
			}
		}
	}
	
	static class node{
		int x;
		int step;//轉換步數
		node(){}
		node(int a,int b){
			x=a;
			step=b;
		}
	}
	public static int BFS(int x,int y) {
		for(int i=0;i<maxn;i++) {//初始化
			vis[i]=false;
		}
		
		int a[]=new int[4];
		Queue<node> q=new LinkedList<node>();
		q.offer(new node(x,0));
		vis[x]=true;
		while(!q.isEmpty()) {
			node now=q.poll();
			if(now.x==y) {
				return now.step;
			}
			x=now.x;
			a[0] = x / 1000;
	        a[1] = (x / 100) % 10;
	        a[2] = (x / 10) % 10;
	        a[3] = x % 10;
	        for (int j = 0; j < 4;j++){
	            int num = a[j];
	            for (int i = 0; i < 10;i++){
	            	if(i!=num){
	                   a[j] = i;
	                   int sum = a[0] * 1000 + a[1] * 100 + a[2] * 10 + a[3];
	                   if(!vis[sum]&&isprim[sum]){
	                	   //System.out.println(sum);
	                       vis[sum] = true;
	                       q.offer(new node(sum,now.step+1));
	                   }
	               }
	           }
	           a[j] = num;
	       }
		}
		return -1;
	}
	public static void main(String[] args) {
		Era(maxn-1);
		Scanner sc=new Scanner(System.in);
		int n,a,b;
		n=sc.nextInt();
		while((n--)>0) {
			a=sc.nextInt();
			b=sc.nextInt();
			int ans=BFS(a,b);
			if(ans==-1) {
				System.out.println("Impossible");
			}
			else {
				System.out.println(ans);
			}
		}
	}
}