Prime Path （bfs+优化）

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

31033 81791373 80171033 1033

Sample Output

670

改了好几次终于对了，无论多么乱，多么长，代码还是要清晰，对于素数的判断，用的多的话，可以直接用一个数组来直接存储嘛，这里的素数存储很巧妙，主要来对合数进行处理，未处理的就是素数，对于每一个因子算式进行枚举

#include <iostream>#include <queue>#include <cstdio>#include <cstring>#include <cmath>#define maxx 10010using namespace std;int n,m;int visit[maxx];int isprime[maxx];struct note{    int x,step;}st,ed;//相对快速的素数标记方法void judge_prime(){    int i,j;    memset(isprime,0,sizeof(isprime));    isprime[0]=1;isprime[1]=1;    for(i=2;i<(sqrt(maxx)+1);i++)    {        if(!isprime[i])        {            for(j=i*i;j<=maxx;j=j+i)            {                isprime[j]=1;            }        }    }}int bfs(){    int i;    queue<note>Q;    st.x=n;    st.step=0;    Q.push(st);    visit[n]=1;    while(!Q.empty())    {        st=Q.front();        Q.pop();        if(st.x==m)        {            return st.step;        }        int k=st.x/1000;        for(i=1;i<k;i++)        {            ed.x=st.x-i*1000;            if(!isprime[ed.x]&&!visit[ed.x])            {                ed.step=st.step+1;                Q.push(ed);                visit[ed.x]=1;            }        }        for(i=1;i<(10-k);i++)        {            ed.x=st.x+i*1000;            if(!isprime[ed.x]&&!visit[ed.x])            {                ed.step=st.step+1;                Q.push(ed);                visit[ed.x]=1;            }        }        k=(st.x-k*1000)/100;        for(i=1;i<=k;i++)        {            ed.x=st.x-i*100;            if(!isprime[ed.x]&&!visit[ed.x])            {                ed.step=st.step+1;                Q.push(ed);                visit[ed.x]=1;            }        }        for(i=1;i<(10-k);i++)        {            ed.x=st.x+i*100;            if(!isprime[ed.x]&&!visit[ed.x])            {                ed.step=st.step+1;                Q.push(ed);                visit[ed.x]=1;            }        }        k=(st.x/10)%10;        for(i=1;i<=k;i++)        {            ed.x=st.x-i*10;            if(!isprime[ed.x]&&!visit[ed.x])            {                ed.step=st.step+1;                Q.push(ed);                visit[ed.x]=1;            }        }        for(i=1;i<(10-k);i++)        {            ed.x=st.x+i*10;            if(!isprime[ed.x]&&!visit[ed.x])            {                ed.step=st.step+1;                Q.push(ed);                visit[ed.x]=1;            }        }        k=st.x%10;        for(i=2;i<=k;i=i+2)        {            ed.x=st.x-i;            if(!isprime[ed.x]&&!visit[ed.x])            {                ed.step=st.step+1;                Q.push(ed);                visit[ed.x]=1;            }        }        for(i=2;i<(10-k);i=i+2)        {            ed.x=st.x+i;            if(!isprime[ed.x]&&!visit[ed.x])            {                ed.step=st.step+1;                Q.push(ed);                visit[ed.x]=1;            }        }    }    return 0;}int main(){    int num;    judge_prime();    scanf("%d",&num);    while(num--)    {        scanf("%d %d",&n,&m);        memset(visit,0,sizeof(visit));        if(n==m)        {            printf("0\n");            continue ;        }        else        {            int ans=bfs();            if(ans==0)                printf("Impossible\n");            else                printf("%d\n",ans);        }    }    return 0;}