prime path poj 3126

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).

One line for each case, either with a number stating the minimal cost or containing the word Impossible.

#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;
# define r 11311
int b[11111];
struct $
{
    int e,s;
} q,z;
void eilish()
{
    memset(b,0,sizeof b);
    b[1]=1;
    for(int i=2; i<r; i++)
    {
        if(!b[i])
        {
            for(int j=2; j*i<r; j++)
                b[i*j]=1;
        }
    }
}
int by(int f,int l)
{
    int v[r];
    int t;
    memset(v,0,sizeof v);
    queue<$>p;
    q.s=0,q.e=f;
    v[f]=1;
    p.push(q);
    while(!p.empty())
    {
        q=p.front();
        p.pop();
        if(q.e==l)
            return q.s;
        int a[5];
        a[1]=q.e/1000;
        a[2]=q.e/100%10;
        a[3]=q.e/10%10;
        a[4]=q.e%10;
        for(int i=1; i<=4; i++)
        {
            int m=a[i];
            for(int j=0; j<=9; j++)
            {
                if(a[i]!=j)
                {
                    a[i]=j;
                    t=a[1]*1000+a[2]*100+a[3]*10+a[4];
                }
                if(t>=1000&&t<=9999&&!v[t]&&!b[t])
                {
                    z.e=t,z.s=q.s+1;
                    p.push(z);
                    v[t]=1;
                }
            }
            a[i]=m;
        }
    }
    return -1;
}
int main()
{
    int n,g,h;
    eilish();
    scanf("%d",&n);
    while(n--)
    {
        scanf("%d%d",&g,&h);
        if(by(g,h)==-1)
            printf("Impossible\n");
        else
            printf("%d\n",by(g,h));
    }
    return 0;
}

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