POJ - 3126 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 Impossible.
Sample Input
3
1033 8179
1373 8017
1033 1033
Sample Output
6
7
0
将一个素数通过变换一位，变换后还是素数，最少几次可以变换到另一个素数。
先打表记录素数，然后将第一个数入队，找到新的素数，加入队列，继续搜索。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>

using namespace std;

int vis[10005];
int prime[10005];

struct node
{
    int num,step;
};

void is_prime()
{
    int i,j;
    memset(prime,0,sizeof(prime));
    for(i=3;i<=100;i+=2)
    {
        if(prime[i]==0)
        {
            for(j=i*i;j<10005;j+=2*i)
            prime[j]=1;
        }
    }
}

int bfs(int a,int b)
{
    queue<node> q;
    node s1,s2;
    int x1,x2,x3,x4,y;
    s1.num=a;
    s1.step=0;
    vis[a]=1;
    q.push(s1);
    while(!q.empty())
    {
        s1=q.front();
        q.pop();
        if(s1.num==b)
        {
            return s1.step;
        }
        for(int i=0;i<4;i++)
        {
            for(int j=0;j<10;j++)
            {
                if(i==0&&j==0)continue;//千位不能为0
                x1=s1.num/1000;
                x2=(s1.num/100)%10;
                x3=(s1.num/10)%10;
                x4=s1.num%10;
                if(i==0)
                {
                    y=j*1000+x2*100+x3*10+x4;
                }
                if(i==1)
                {
                    y=x1*1000+j*100+x3*10+x4;
                }
                if(i==2)
                {
                    y=x1*1000+x2*100+j*10+x4;
                }
                if(i==3)
                {
                    y=x1*1000+x2*100+x3*10+j;
                }
                if(y!=s1.num&&!vis[y]&&!prime[y]&&y%2==1)//判断y是否为素数，是否访问过，不能为偶数，偶数这点我确实没想到。
                {
                    vis[y]=1;
                    s2.num=y;
                    s2.step=s1.step+1;
                    q.push(s2);
                }
            }
        }
    }
    return -1;
}

int main()
{
    int t;
    scanf("%d",&t);
    is_prime();
    while(t--)
    {
        int a,b;
        scanf("%d%d",&a,&b);
        memset(vis,0,sizeof(vis));
        int ans=bfs(a,b);
        if(ans!=-1)
        {
            printf("%d\n",ans);
        }
        else
        {
            printf("Impossible\n");
        }
    }
    return 0;
}