Poj 3126 Prime Path

1.Link:

http://poj.org/problem?id=3126

2.Content:

Prime Path

Time Limit: 1000MS Memory Limit: 65536KTotal 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

31033 81791373 80171033 1033

Sample Output

670

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\n",i);31        else printf("%d不是素数\n",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             else46             {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: