Prime Path（BFS）

Prime Path

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other)

Total Submission(s) : 21   Accepted Submission(s) : 18

Problem 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

PKU

 

题意：

给出两个四位的素数，要求求出从其中一个变化到另一个数的最少的变化次数，每一次变化只变化四位中的一位，并且变化后的数也要是素数；

思路：

bfs，只不过是40入口的bfs，需要经过剪枝；每一次都枚举个位、十位、百位、千位的所有变化，检验室素数后加入到队列中；

AC代码：

 #include<iostream>
 #include<cstdio>
 #include<cstring>

 using namespace std;
 int a,b;
 struct kf
 {
     int number;
     int sgin;
 }ks[];
 bool ksgin[]={false};

 bool shu(int sg)//判断sg是否是素数
 {
     if(sg==||sg==)
         return true;
     else if(sg<=||sg%==)
         return false;
     else if(sg>)
     {
         for(int i=;i*i<=sg;i+=)
             if(sg%i==)
                 return false;
         return true;
     }
 }

 int bfs()
 {
     int left,right;
     kf s;
     ks[left=right=].number=a;
     ks[right++].sgin=;
     ksgin[a]=false;
     while(left<right){
         s=ks[left++];
         if(s.number==b){
             cout<<s.sgin<<endl;
             return ;
         }
         int ge=s.number%;
         int shi=(s.number/)%;
         for(int i=;i<=;i+=){//枚举个位
             int y=s.number/*+i;
             if(y!=s.number&&ksgin[y]&&shu(y)){
                 ksgin[y]=false;
                 ks[right].number=y;
                 ks[right++].sgin=s.sgin+;
             }
         }
         for(int i=;i<=;i++){//枚举十位
             int y=s.number/*+i*+ge;
             if(y!=s.number&&ksgin[y]&&shu(y)){
                 ksgin[y]=false;
                 ks[right].number=y;
                 ks[right++].sgin=s.sgin+;
             }
         }
         shi*=;
         shi+=ge;
         for(int i=;i<=;i++){//枚举百位
             int y=s.number/*+i*+shi;
             if(y!=s.number&&ksgin[y]&&shu(y)){
                 ksgin[y]=false;
                 ks[right].number=y;
                 ks[right++].sgin=s.sgin+;
             }
         }
         shi=s.number%;
         for(int i=;i<=;i++){//千位
             int y=i*+shi;
             if(y!=s.number&&ksgin[y]&&shu(y)){
                 ksgin[y]=false;
                 ks[right].number=y;
                 ks[right++].sgin=s.sgin+;
             }
         }
     }
     cout<<"Impossible"<<endl;
     return ;
 }

 int main()
 {
 //    freopen("input.txt","r",stdin);
     int test;
     cin>>test;
     while(test--){
         memset(ksgin,true,sizeof(ksgin));
         cin>>a>>b;
         bfs();
     }
 }