poj 3126 Prime Path

Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 16198 Accepted: 9148
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
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
Source
Northwestern Europe 2006
题目大意：
给定两个四位素数a b，要求把a变换到b
变换的过程要保证 每次变换出来的数都是一个 四位素数，而且当前这步的变换所得的素数 与 前一步得到的素数 只能有一个位不同，而且每步得到的素数都不能重复。
求从a到b最少需要的变换次数。无法变换则输出Impossible

/*
埃式筛法素数判定+BFS+可行解判定+乱搞.
我们先搞出[1000,10000]的prime.
然后带入prime判是否可行（只有一位不同）
用BFS跑最优解（神奇的BFS）.
*/
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#define MAXN 10001
using namespace std;
int n,m,t,s[MAXN],tot,head,tail;
bool b[MAXN];
struct data
{
    int step;
    int v;
}
q[MAXN];
void prime()
{
    for(int i=2;i<=10000;i++)
      if(!b[i])
      {
        for(int j=i+i;j<=10000;j+=i)
            b[j]=true;
      }
    for(int i=1001;i<=10000;i++)
      if(!b[i])
          s[++tot]=i;
}
bool jd(int n,int m)
{
    int sum=0;
    while(n)
    {
        int i=n%10;n/=10;
        int j=m%10;m/=10;
        if(i!=j) sum++;
    }
    if(sum==1) return true;
    return false;
}
bool bfs(int n,int m)
{
    head=0,tail=1;
    q[tail].step=0;
    q[tail].v=n;
    b[n]=true;
    while(head<tail)
    {
        head++;
        if(q[head].v==m) return true;
        for(int i=tot;i>=1;i--)
        {
            if(!b[s[i]]&&jd(q[head].v,s[i]))
            {
                b[s[i]]=true;
                q[++tail].v=s[i];
                q[tail].step=q[head].step+1;
            }
        }
    }
    return false;
}
int main()
{
    prime();
    cin>>t;
    while(t--)
    {
        memset(b,0,sizeof(b));
        scanf("%d %d",&n,&m);
        if(bfs(n,m))
        {
            printf("%d
",q[head].step);
        }
        else  printf("Impossible
");
    }
    return 0;
}