G - Prime Path

题目：

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

题意：

一个四位数，每次改变个十百千位的其中一位，使之变成一个四位素数，求由a到b进行变化的最小的次数；

思路：

首先要进行打表，求出所有的素数，否则时间可能会超限，然后再用广搜进行搜索；

代码如下：

#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;

int t,aa,bb;
int s[11000];

void init()//素数进行打表，
{
    int i,j;
    memset(s,0,sizeof s);
    s[1]=1;
    for(i=2;i<11000;i++)
    {
        if(s[i]==0)//是素数
        {
            for(j=2;j*i<11000;j++)
            {
                s[i*j]=1;
            }
        }
    }
    return ;
}

struct node
{
    int x;
    int step;
};
int book[11000];

int bfs()
{
    int i,j,w,e;
    int q[5];
    node a,next;
    queue<node>Q;
    memset(book,0,sizeof book);
    a.x=aa;
    a.step=0;
    book[aa]=1;
    Q.push(a);
    while(!Q.empty())
    {
        a=Q.front();
        Q.pop();
        if(a.x==bb)
            return a.step;
        q[0]=a.x/1000;//取出这个数的千 百 十 个 位；
        q[1]=a.x/100%10;
        q[2]=a.x/10%10;
        q[3]=a.x%10;
        for(i=0;i<4;i++)
        {
            e=q[i];//进行记录，保持原来的数值不变；
            for(j=0;j<10;j++)
            {
                if(q[i]!=j)
                {
                    q[i]=j;
                    w=q[0]*1000+q[1]*100+q[2]*10+q[3];
                    if(w<1000||w>10000||book[w]==1||s[w]==1)//不是四位数，搜索过，不是素数，continue；
                        continue;
                    next.x=w;
                    next.step=a.step+1;
                    Q.push(next);
                    book[w]=1;
                }
            }
            q[i]=e;//恢复；
        }
    }
    return -1;
}

int main()
{
    scanf("%d",&t);
    init();
    while(t--)
    {
        int i,j;
        scanf("%d%d",&aa,&bb);
        int kk=bfs();
        if(kk==-1)
            printf("Impossible\n");
        else
            printf("%d\n",kk);
    }
    return 0;
}