Prime Path(bfs最短步数问题)

Prime Path

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 18362		Accepted: 10327

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

题意：给定两个素数n和m，要求把n变成m，每次变换时只能变一个数字，即变换后的数与变换前的数只有一个数字不同，并且要保证变换后的四位数也是素数。求最小的变换次数；如果不能完成变换，输出Impossible。

无论怎么变换，个位数字一定是奇数（个位数字为偶数肯定不是素数），这样枚举个位数字时只需枚举奇数就行；而且千位数字不能是0。所以可以用广搜，枚举各个数位上的数字，满足要求的数就加入队列，直到变换成功。因为是广搜，所以一定能保证次数最少。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <cmath>
using namespace std;
struct node
{
  int step;
  int x;
}s,v;
int n,m;
int vis[10000];
int judge(int x)
{
   if(x==0||x==1)
   return 0;
   else if(x==2||x==3)
   return 1;
   else
   {
      for(int i=2;i<=(int)sqrt(x);i++)
      {
          if(x%i==0)
          return 0;
      }
      return 1;
   }
}
void bfs()
{
   queue<struct node>q;
   s.x=n;
   s.step=0;
   q.push(s);
   while(!q.empty())
   {
      v=q.front();
      q.pop();
      if(v.x==m)
      {
         printf("%d\n",v.step);
         return;
      }
      for(int i=1;i<=9;i+=2)//处理个位
      {
          int num=v.x/10*10+i;
          if(!vis[num]&&judge(num)&&num!=v.x)
          {
             vis[num]=1;
             s.x=num;
             s.step=v.step+1;
             q.push(s);
          }
      }
      for(int i=0;i<=9;i++)//处理十位
      {
          int num=v.x/100*100+i*10+v.x%10;
          if(!vis[num]&&judge(num)&&num!=v.x)
          {  vis[num]=1;
             s.x=num;
             s.step=v.step+1;
             q.push(s);
          }
      }
      for(int i=0;i<=9;i++)处理百位
      {
          int num=v.x/1000*1000+i*100+v.x%100;
          if(!vis[num]&&judge(num)&&num!=v.x)
          {  vis[num]=1;
             s.x=num;
             s.step=v.step+1;
             q.push(s);
          }
      }
      for(int i=1;i<=9;i++)处理千位
      {
          int num=i*1000+v.x%1000;
          if(!vis[num]&&judge(num)&&num!=v.x)
          {  vis[num]=1;
             s.x=num;
             s.step=v.step+1;
             q.push(s);
          }
      }
   }
   printf("Impossible\n");
   return;
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
       scanf("%d%d",&n,&m);
       memset(vis,0,sizeof(vis));
       vis[n]=1;
       bfs();
    }
    return 0;
}