Poj 3126 Prime Path

1.Link:

2.Content:

Prime Path

Time Limit: 1000MS Memory Limit: 65536K

Total Submissions: 11757 Accepted: 6675

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

3.Method:

4.Code:

 1 #include<iostream>
 2 #include<queue>
 3 using namespace std;
 4 #define MAX 10002
 5 //#define MAX 102
 6 int a[MAX];
 7 int b[MAX];
 8 int main()
 9 {
10     int n,m;
11     int i,j;
12     int x,y;
13     int k;
14     queue<int> q;
15     a[2]=0; 
16     for(i=4;i<MAX;i+=2) a[i]=1; 
17     for(i=3;i<=MAX/2;i++)
18     {
19         if(a[i]==0)
20         {
21            for(j=i;j<MAX/i;j+=2)
22            {
23                a[i*j]=1;     
24            }
25         }
26     }
27     //验证素数表的正确性
28     /*for(i=2;i<MAX;i++)
29     {
30        if(a[i]==1) printf("%d是素数
",i);
31        else printf("%d不是素数
",i);
32     }*/ 
33     cin>>n;
34     for(int ii=0;ii<n;ii++)
35     {
36         cin>>x>>y;
37         for(j=0;j<MAX;j++) b[j]=-1;
38         while(!q.empty()) q.pop();
39         q.push(x);
40         b[x]=0;
41         while(!q.empty())
42         {
43             m=q.front();
44             if(m==y) break;
45             else
46             {
47                 for(i=0;i<4;i++)
48                 {
49                     k=1;
50                     for(j=0;j<i;j++) k=k*10;
51                     for(j=-9;j<=9;j++)
52                     {
53                         if(((m/(k*10))==((m+j*k)/(k*10)))&&(a[m+j*k]==0)&&(b[m+j*k]==-1)&&(m+j*k)>=1000)
54                         {
55                              //printf("perfect");
56                              b[m+j*k]=b[m]+1;
57                              q.push(m+j*k);
58                         }
59                     }
60                 }
61             }
62             q.pop();
63         }
64         cout<<b[y]<<endl;   
65     }
66     //system("pause");
67     return 1;
68 }

5.Reference:

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 11757		Accepted: 6675