i-09

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

31033 81791373 80171033 1033

Sample Output

670

题意：
四位数，每次改变一位上的数，使其变成素数，直到达到要求，求改变的步数

分析：

用宽搜就行

代码：

[cpp] view plain copy

1. #include<iostream>  
2. #include<cstring>  
3. #include<queue>  
4. #include<algorithm>  
5. using namespace std;  
6. int a[10000],b[10000];  
7. struct num  
8. {  
9.     int s,l;  
10. };  
11. void bfs(int n,int m)  
12. {  
13.     num v,kk;  
14.     int i,sum=0,temp;  
15.     v.s=n;  
16.     v.l=0;  
17.     queue<num>gg;  
18.     gg.push(v);  
19.     while(!gg.empty())  
20.     {  
21.         v=gg.front();  
22.         if(v.s==m){cout<<v.l<<endl;return ;}  
23.         gg.pop();  
24.         for(i=1;i<10;i++)  
25.         {  
26.             temp=v.s%1000+i*1000;  
27.             if(b[temp]==0&&!a[temp])  
28.             {  
29.                 kk.s=temp;  
30.                 kk.l=v.l+1;  
31.                 b[temp]=1;  
32.                 gg.push(kk);  
33.             }  
34.         }  
35.         for(i=0;i<10;i++)  
36.         {  
37.             temp=v.s/1000*1000+v.s%100+i*100;  
38.             if(b[temp]==0&&!a[temp])  
39.             {  
40.                 kk.s=temp;  
41.                 kk.l=v.l+1;  
42.                 b[temp]=1;  
43.                 gg.push(kk);  
44.             }  
45.         }  
46.         for(i=0;i<10;i++)  
47.         {  
48.             temp=v.s/100*100+v.s%10+i*10;  
49.             if(b[temp]==0&&!a[temp])  
50.             {  
51.                 kk.s=temp;  
52.                 kk.l=v.l+1;  
53.                 b[temp]=1;  
54.                 gg.push(kk);  
55.             }  
56.         }  
57.         for(i=0;i<10;i++)  
58.         {  
59.             temp=v.s/10*10+i;  
60.             if(b[temp]==0&&!a[temp])  
61.             {  
62.                 kk.s=temp;  
63.                 kk.l=v.l+1;  
64.                 b[temp]=1;  
65.                 gg.push(kk);  
66.             }  
67.         }  
68.     }  
69.     cout<<"Impossible"<<endl;  
70.     return;  
71. }  
72. int main()  
73. {  
74.     int t,k,n,m,i,j;  
75.     for(i=2;i<10000;i++)  
76.     {  
77.         if(a[i]==0)  
78.         {  
79.             for(j=i+1;j<10000;j++)  
80.             {  
81.                 if(j%i==0)a[j]=1;  
82.             }  
83.         }  
84.     }  
85.     cin>>t;  
86.     for(k=0;k<t;k++)  
87.     {  
88.         memset(b,0,sizeof(b));  
89.         cin>>n>>m;  
90.         bfs(n,m);  
91.     }  
92. }  

感受：

挺水的