Poj 3126 Prime Path

1.Link:

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

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:

#include<iostream>
#include<queue>
using namespace std;
#define MAX 10002
//#define MAX 102
int a[MAX];
int b[MAX];
int main()
{
    int n,m;
    int i,j;
    int x,y;
    int k;
    queue<int> q;
    a[2]=0; 
    for(i=4;i<MAX;i+=2) a[i]=1; 
    for(i=3;i<=MAX/2;i++)
    {
        if(a[i]==0)
        {
           for(j=i;j<MAX/i;j+=2)
           {
               a[i*j]=1;     
           }
        }
    }
    //验证素数表的正确性
    /*for(i=2;i<MAX;i++)
    {
       if(a[i]==1) printf("%d是素数
",i);
       else printf("%d不是素数
",i);
    }*/ 
    cin>>n;
    for(int ii=0;ii<n;ii++)
    {
        cin>>x>>y;
        for(j=0;j<MAX;j++) b[j]=-1;
        while(!q.empty()) q.pop();
        q.push(x);
        b[x]=0;
        while(!q.empty())
        {
            m=q.front();
            if(m==y) break;
            else
            {
                for(i=0;i<4;i++)
                {
                    k=1;
                    for(j=0;j<i;j++) k=k*10;
                    for(j=-9;j<=9;j++)
                    {
                        if(((m/(k*10))==((m+j*k)/(k*10)))&&(a[m+j*k]==0)&&(b[m+j*k]==-1)&&(m+j*k)>=1000)
                        {
                             //printf("perfect");
                             b[m+j*k]=b[m]+1;
                             q.push(m+j*k);
                        }
                    }
                }
            }
            q.pop();
        }
        cout<<b[y]<<endl;   
    }
    //system("pause");
    return 1;
}

5.Reference: