poj 3126 Prime Path(BFS)

Prime Path

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7574		Accepted: 4302

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

Source

Northwestern Europe 2006

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#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
#include <cmath>
using namespace std;
struct node
{
    int val,dis;
};
bool is(int &n)
{
    if(n%2==0) return false;
    int i,m=sqrt(double(n));
    for(i=3;i<=m;i++)
     if(n%i==0)
      return false;
    return true;
}
int n,m;
int Cos(int val)
{
  bool h[10000]={0};
  int i,k,t;
  node a,b;a.dis=0;a.val=val;
  queue<node >Q;
  h[val]=1;
  Q.push(a);
  while(!Q.empty())
  {
    a=Q.front();Q.pop();
    k=a.val/10*10;
    for(i=1;i<=9;i+=2)
    {
        t=k+i;
       if(is(t)&&!h[t])
       {
           if(t==m)
            return a.dis+1;
         h[t]=1;
         b.val=t;
         b.dis=a.dis+1;
         Q.push(b);
       }
    }
    t=a.val%10;
    k=a.val/100*100+t;
    for(i=0;i<=9;i++)
    {
        t=k+i*10;
       if(is(t)&&!h[t])
       {
         if(t==m)
            return a.dis+1;
         h[t]=1;
         b.val=t;
         b.dis=a.dis+1;
         Q.push(b);
       }
    }
    t=a.val%100;
    k=a.val/1000*1000+t;
    for(i=0;i<=9;i++)
    {
        t=k+i*100;
       if(is(t)&&!h[t])
       {
           if(t==m)
            return a.dis+1;
         h[t]=1;
         b.val=t;
         b.dis=a.dis+1;
         Q.push(b);
       }
    }
    k=a.val%1000;
    for(i=1;i<=9;i++)
    {
        t=k+i*1000;
       if(is(t)&&!h[t])
       {
           if(t==m)
            return a.dis+1;
         h[t]=1;
         b.val=t;
         b.dis=a.dis+1;
         Q.push(b);
       }
    }
  }
  return -1;
}
int main()
{
   int t;
   scanf("%d",&t);
   int cost;
   while(t--)
   {
       scanf("%d%d",&n,&m);
       if(n==m)  {printf("0\n");continue;}
       cost=Cos(n);
       if(cost==-1)
        printf("Impossible\n");
      else
        printf("%d\n",cost);
   }
    return 0;
}