POJ 3126 Prime Path

Prime Path

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other)

Total Submission(s) : 7   Accepted Submission(s) : 3

Problem 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

31033 81791373 80171033 1033

 

Sample Output

670

 

Source

PKU

 

一道简单的BFS+素数读表：

#include <iostream>
#include <cstring>
#include <cmath>
#include <queue>
 
using namespace std;
 
int isprime[10005];
 
struct nnd
{
    int first;
    int second;
};
 
int a,e;
int vis[10005];
 
void dfs(queue<nnd> q)
{
    while(!q.empty())
    {
        nnd t;
        nnd r;
        t=q.front();
        q.pop();
        //cout<<e<<endl;
        if(t.first==e)
        {
            //cout<<"hello";
            cout<<t.second<<endl;
            return ;
        }
 
        int k=t.first%1000;
        for(int i=1000;i<10000;i=i+1000)
        {
            if(isprime[k+i]&&!vis[k+i])
            {
               r.first=k+i;   r.second=t.second+1;
               q.push(r);
               vis[i+k]=1;
            }
        }
 
        int kk=t.first%100;
        int kkk=t.first/1000*1000;
        k=kk+kkk;
        for(int i=0;i<1000;i=i+100)
        {
            if(isprime[i+k]&&!vis[i+k])
            {
                r.first=i+k;
                r.second=t.second+1;
                q.push(r);
                vis[i+k]=1;
            }
        }
 
        kk=t.first%10;
        kkk=t.first/100*100;
        k=kk+kkk;
        for(int i=0;i<100;i=i+10)
        {
            if(isprime[i+k]&&!vis[i+k])
            {
                r.first=i+k;
                r.second=t.second+1;
                q.push(r);
                vis[i+k]=1;
            }
        }
 
        k=t.first/10*10;
        for(int i=1;i<10;i=i+2)
        {
            if(isprime[i+k]&&!vis[i+k])
            {
                r.first=i+k;
                r.second=t.second+1;
                q.push(r);
                vis[i+k]=1;
            }
        }
    }
 
    cout<<"Impossible"<<endl;
    return;
}
 
 
int main()
{
    int n;
    cin>>n;
 
    for(int i=0;i<10005;i++)
    {
        isprime[i]=i;
    }
    for(int i=2;i*i<10010;i++)
    {
        if(isprime[i]!=0)
        for(int j=2;j*i<10010;j++)
        {
            isprime[i*j]=0;
        }
    }
 
    isprime[1]=0;
 
 
    for(int o=0;o<n;o++)
    {
        queue<nnd> q;
        nnd a;
        memset(vis,0,sizeof(vis));
        cin>>a.first>>e;
        a.second=0;
        q.push(a);
        vis[a.first]=1;
 
        dfs(q);
    }
    return 0;
}