Prime Path（POJ 3126 BFS）

Prime Path

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 15325		Accepted: 8634

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
单纯的将所有情况都搜一遍

#include <iostream>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
bool prime[10000];
bool vis[10000];
int n[15];
int a,b;
struct node
{
    int num,t;
};
void init()
{
    int i,j;
    memset(prime,0,sizeof(prime));
    for(i=1000;i<9999;i++)
    {
        for(j=2;j<i;j++)
        {
            if(i%j==0)
                break;
        }
        if(j==i)
        {
            prime[i]=1;    //1代表质数
            //cout<<i<<endl;
        }
    }
    for(i=0;i<10;i++)
        n[i]=i;
    n[10]=1,n[11]=3,n[12]=7,n[13]=9;
    return;
}
int bfs()
{
    int i,j;
    queue<node> Q;
    node tem,u;
    int k;
    tem.num=a,tem.t=0;
    memset(vis,0,sizeof(vis));
    Q.push(tem);
    while(!Q.empty())
    {
        tem=Q.front();
        Q.pop();
        //cout<<tem.num<<endl;
        /*if(tem.num==3733)
            printf("adsfadsf
");*/
        if(tem.num==b)
            return tem.t;
        for(i=10;i<=13;i++)
        {
            if(tem.num%10==n[i])
                continue;
            u.num=tem.num/10;
            u.num=u.num*10+n[i];
            //cout<<u.num<<" "<<prime[u.num]<<" "<<vis[u.num]<<endl;
            if(vis[u.num]==0&&prime[u.num])
            {
                u.t=tem.t+1;
                vis[u.num]=1;
                Q.push(u);
            }
        }
        for(i=0;i<=9;i++)
        {
            k=(tem.num/10)%10;
            if(k==n[i])
                continue;
            k=tem.num%10;
            u.num=((tem.num/100)*10+n[i])*10+k;
            if(vis[u.num]==0&&prime[u.num])
            {
                u.t=tem.t+1;
                vis[u.num]=1;
                Q.push(u);
            }
        }
        for(i=0;i<=9;i++)
        {
            k=(tem.num/100)%10;
            if(k==n[i])
                continue;
            k=tem.num%100;
            u.num=((tem.num/1000)*10+n[i])*100+k;
            if(vis[u.num]==0&&prime[u.num])
            {
                u.t=tem.t+1;
                vis[u.num]=1;
                Q.push(u);
            }
        }
        for(i=1;i<=9;i++)
        {
            k=tem.num/1000;
            if(k==n[i])
                continue;
            //cout<<k<<" "<<n[i]<<endl;
            k=tem.num%1000;
            u.num=n[i]*1000+k;
        //    cout<<u.num<<endl;
            if(vis[u.num]==0&&prime[u.num])
            {
                u.t=tem.t+1;
                vis[u.num]=1;
                //cout<<u.num<<endl;
                Q.push(u);
            }
        }
        //break;
    }
    return -1;
}    
int main()
{
    int T,ans;
    init();
    freopen("in.txt","r",stdin);
    //freopen("ou.txt","w",stdin);
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&a,&b);
        ans=bfs();
        if(ans==-1)    printf("Impossible
");
        else printf("%d
",ans);
    }
}