poj 3126 BFS

Prime Path

Time Limit: 1000MS
Memory Limit:65536K

Total Submissions: 6713
Accepted:3808

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

#include <iostream>
#include <queue>
#include <cstring>
#include <cmath>
#include <cstdio>
using namespace std;
int first,last;
int step[10000];
bool isPrime(int a)
{
    bool flag=true;
    for(int i=2;i<=(sqrt(float(a)));++i)
        if(a%i==0){flag=false;break;}
    return flag;
}
int bfs()
{
    memset(step,-1,sizeof(step));
    queue<int> s;
    s.push(first);
    step[first] = 0;
    int temp;
    while(!s.empty())
    {
        int dequeue = s.front();
        s.pop();
        if(dequeue == last)
            return step[dequeue];
        for(int i=0;i<10;++i)
        {
            temp = (dequeue/10)*10+i;
            if(isPrime(temp) && step[temp] == -1)
            {
                step[temp] = step[dequeue] + 1;
                s.push(temp);
            }
        }
        for(int i=0;i<10;++i)
        {
            temp = (dequeue/100)*100+dequeue%10+i*10;
            if(isPrime(temp) && step[temp] == -1)
            {
                step[temp] = step[dequeue] + 1;
                s.push(temp);
            }
        }
        for(int i=0;i<10;++i)
        {
            temp = (dequeue/1000)*1000+dequeue%100+i*100;
            if(isPrime(temp) && step[temp] == -1)
            {
                step[temp] = step[dequeue] + 1;
                s.push(temp);
            }
        }
        for(int i=1;i<10;++i)
        {
            temp = dequeue%1000+i*1000;
            if(isPrime(temp) && step[temp] == -1)
            {
                step[temp] = step[dequeue] + 1;
                s.push(temp);
            }
        }

    }
    return -1;
}
int main()
{
    int n,result;
    scanf("%d",&n);
    while(n--)
    {
        scanf("%d%d",&first,&last);
        result = bfs();
        if(result == -1) printf("Impossible\n");
        else printf("%d\n",result);
    }
    return 0;
}