Prime Path (POJ

题目描述：
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
大致意思是给你两个素数x和y，且都是4位数
x变换到y的步骤是每次只改变x的一个数字，且这个过度状态下的数也必须满足是4位的素质，问最少需要几步。

用宽搜的两个条件是，一个是存在路径，二十求的是最值或者是打印路径。
所以大致确定是用宽搜，即使这道题打印路径也不会很难
那么我们需要确定一个状态（题目里是表示一个素数）如何转移到另一个状态，这个不难，就比如 初始状态1033，对第一位有9种其他的数可以替换，其他位也是，所以总共有4*9=36种状态转移方向，但是是否满足一个合法转移，需要两个条件，这个数是4位素数，并且在这个前提下，这个素数之前并没有访问过。

问题不是很难，但是需要耐心，代码打着打着就非常长了，（上面还有很多细节没有涉及，如何判断素数，这需要打素数表）
代码：

import java.util.ArrayList;import java.util.HashSet;import java.util.LinkedList;import java.util.Queue;import java.util.Scanner;public class Main{       static int vv[];    public static void main(String[] args)     {        ArrayList<Integer> list=new ArrayList<Integer>();        HashSet<Integer>set=new HashSet<Integer>();        int num=2;        int p=1;        boolean sign;        while(num<9999)//找出10000以内的素数        {            if(p*p<num)p++;            sign=true;            for(int i=0;i<list.size()&&list.get(i)<=p;i++)            {                if(num%list.get(i)==0)                {                    sign=false;                    break;                }            }            if(sign)            {                list.add(num);                if(num>=1000)//我们只要大于1000的素数，加入hax表用于判断                    set.add(num);            }            num++;        }//前期工作完成        Scanner sc=new Scanner(System.in);        int n=sc.nextInt();        while((n--)>0)        {            int x=sc.nextInt();//开始状态            int y=sc.nextInt();//结束状态            if(x==y)//相等直接输出            {                System.out.println(0);                continue;            }            vv=new int[10000];//判断这个数（状态）是否访问过，置1说明访问过了            Queue<node>queue =new LinkedList<node>();            queue.add(new node(x,0));//初始状态            vv[x]=1;            int ans=0;            boolean sign1=true;//为true表示没有找到            while(!queue.isEmpty()&&sign1)            {                node t=queue.poll();//出队                int k=t.s;                int a[]=new int[4];                int g=1000;                for(int i=0;i<=3;i++)//这一步是把这个素质分解成各个位上的4个数，以方便下面的枚举                {                    a[i]=k/g;                    k=k%g;                    g/=10;                }                int h;                boolean sign2=false;                for(int i=0;i<=3;i++)//考录每个位数                {                    if(sign2)break;                    h=a[i];//把原先的数保存起来                    for(int j=0;j<=9;j++)//每个位上的数只能0到9，有人说千位上不能为0，没关系，假设这个三位数是素数，这个数也不会出现在set中，然后状态是自己本身怎么办，都没关系，因为下面的if判断绝对通不过（对状态的转移尽量统一化，尽量少管特殊情况，除非万不得已）                    {                        a[i]=j;                        int getT=get(a);//改变了以后转化为一个数                        if(!set.contains(getT)||vv[getT]==1)//判断先如果不在set中，或者在set中，且之前有访问过跳过                            continue;                        if(getT==y)//一旦找到                        {                            sign1=false;//退出while循环                            sign2=true;//退出嵌套的for循环                            ans=t.m+1;//把答案转移到ans                            break;                        }                        else//没有找到，就加入队列，                        {                            queue.add(new node(getT,t.m+1));                            vv[getT]=1;//说明访问过了                        }                    }                    a[i]=h;//，所有可能都枚举完了，还原之前的数                }            }            System.out.println(ans);        }    }    static int get(int a[] )//四个数转化成一个数    {        int s=0;        for(int i=0;i<4;i++)            s=(s*10+a[i]);        return s;    }}class node{    int s;    int m;//在转移到这个状态时，已经走过的步数    node(int s,int m)    {        this.s=s;        this.m=m;    }}