Prime Path（poj3126）

题目：

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

一道简单的bfs宽搜水题，注意先要构造一个素数池即可：

以下是ac代码：

#include<stdio.h>
#include<iostream>
#include<math.h>
using namespace std;
struct node
{
    int x;
    int step;
};
int main()
{
    int prime[10005]={0};//
    for(int i=1000;i<10000;i++)
    {
     //   bool isprime=true;int t=0;
        for(int j=2;j<=sqrt((double)i);j++)
        {
            if(i%j==0)
                prime[i]=1;
        }

    }
    //prime[i]=0;
    //至此 构建成素数池

    int n;
    cin>>n;
    for(int i=0;i<n;i++)
    {
        int x,y,k=0,m=0,step=0;
        cin>>x>>y;
        struct node root,queue[4000];
        int visit[10010]={0};
        root.step=0;root.x=x;
        queue[k++]=root;
        while(true )
        {

            root=queue[m++];
            if(root.x==y)
            {
                step=root.step;
                break;
            }
            else
            {
                for(int i=0;i<40;i++)
                {

                    int j=i/10+1;int t=i%10;
                    struct node q;
                    int a,b,c,d;
                    a=root.x%10;
                    b=root.x/10%10;
                    c=root.x/100%10;
                    d=root.x/1000;
                   // cout<<"&";
                    if(j==1)  q.x=root.x-a+t;
                    else if(j==2)q.x=root.x-b*10+t*10;
                    else if(j==3)q.x=root.x-c*100+t*100;
                    else if(j==4)q.x=root.x-d*1000+t*1000;

                    if(prime[q.x]==1)
                        continue;
                 //       cout<<"&";
                    if(visit[q.x]==1)
                        continue;
                    if(q.x<1000 || q.x>9999)
                        continue;

                    q.step=root.step+1;
                    queue[k++]=q;
                    visit[q.x]=1;

                }
            }

        }
        cout<<step<<endl;
    }
    return 0;
}