Prime Path

 

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.

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.

3

1033 8179

1373 8017

1033 1033

6

7

0

题意  将a转换到b，每次换一位，每次交换后的数都是素数，算最短换了多少次
我的思路
找到所有的四位数的素数，打表
用bfs 有四十个方向，找到答案

 #include <stdio.h>

 #include <stdlib.h>

 #include <string.h>

 int prime[],n,m,tail,head;

 struct que

 {

     int x,time;

 }que[];

 int row(int x)

 {

     int i,s=;

     for(i=;i<=x;i++)

         s*=;

     return s;

 }

 int bfs()

 {

     int i,j,x;

     for(i=;i<=;i++)

     {

         for(j=;j<=i/;j++)

            if(i%j==)break;

         if(j==i/+)prime[i]=;

         else prime[i]=;

     }

     prime[n]=;

     tail=head=;

     que[].x=n,que[].time=;

     while(head<=tail)

     {

         for(i=;i<=;i++)

         {

             for(j=;j<=;j++)

             {

                 x=que[head].x-(que[head].x/row(i-)%)*row(i-)+j*row(i-);

                 if(x>=&&x<=&&prime[x])

                 {

                     que[++tail].x=x;

                     que[tail].time=que[head].time+;

                     if(x==m)return que[tail].time;

                     else  prime[x]=;

                 }

             }

         }

         head++;

     }

     return ;

 }

 int main()

 {

     int i,j,key,g;

     while(scanf("%d",&g)==)

     {

         for(i=;i<g;i++)

         {

             scanf("%d %d",&n,&m);

             if(n==m)

             {

                 printf("0\n");

                 continue;

             }

             key=bfs();

             if(key)

               printf("%d\n",key);

             else

               printf("Impossible\n");

         }

     }

     return ;

 }