Prime Path(素数筛选+bfs)

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9519		Accepted: 5458

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
题意：给定两个四位数，求从前一个数变到后一个数最少需要几步，改变的原则是每次只能改变某一位上的一个数，而且每次改变得到的必须是一个素数；

思路：将四位数以内的素数筛选出来，bfs时，枚举四位数的每一位上的每一个数；

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

struct node
{
    int num;
    int step;
};
queue<struct node>que;
int n,m,flag;
bool p[10010],vis[10010];

//素数筛；
void prime_search()
{
    memset(p,1,sizeof(p));
    for(int i = 4; i <= 10000; i+=2)
        p[i] = 0;
    for(int i = 3; i <= 100; i++)
    {
        if(p[i])
        {
            for(int j = i+i; j <= 10000; j += i)
                p[j] = 0;
        }
    }
}

int bfs()
{
    while(!que.empty())
        que.pop();
    que.push((struct node){n,0});
    vis[n] = 1;
    int res,tmp,r,t,s;
    while(!que.empty())
    {
        struct node u = que.front();
        que.pop();
        if(u.num == m)
            return u.step;
            
        //枚举个位数
        r = u.num%10;
        for(int k = -9; k <= 9; k++)
        {
            t = r+k;
            if(t >= 0 && t <= 9)
            {
                res = (u.num/10)*10+t;
                if(p[res] && !vis[res])
                {
                    que.push((struct node){res,u.step+1});
                    vis[res] = 1;
                }
            }
        }
        
        //枚举十位数
        tmp = u.num/10;
        r = tmp%10;
        s = tmp/10;
        for(int k = -9; k <= 9; k++)
        {
            t = r+k;
            if(t >= 0 && t <= 9)
            {
                res = (s*10+t)*10+u.num%10;
                if(p[res] && !vis[res])
                {
                    que.push((struct node){res,u.step+1});
                    vis[res] = 1;
                }
            }
        }
        
        //枚举百位数
        int tmp = u.num/100;
        r = tmp%10;
        s = tmp/10;
        for(int k = -9; k <= 9; k++)
        {
            t = r+k;
            if(t >= 0 && t <= 9)
            {
                res = (s*10+t)*100+u.num%100;
                if(p[res] && !vis[res])
                {
                    que.push((struct node){res,u.step+1});
                    vis[res] = 1;
                }
            }
        }

        //枚举千位数
        r = u.num/1000;
        for(int k = -9; k <= 9; k++)
        {
            t = r+k;
            if(t >0 && t <= 9)
            {
                res = t*1000+u.num%1000;
                if(p[res] && !vis[res])
                {
                    que.push((struct node){res,u.step+1});
                    vis[res] = 1;
                }
            }
        }
    }
}
int main()
{
    int test;
    prime_search();
    scanf("%d",&test);
    while(test--)
    {
        memset(vis,0,sizeof(vis));
        scanf("%d %d",&n,&m);
        int ans = bfs();
        printf("%d
",ans);
    }
    return 0;
}