Prime Path (poj 3126)（bfs）

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

使用bfs

#include<cstdio>
#include<cstring>
#include<cmath>
#include<queue>

using namespace std;

const int N = 1e4 + 100;

int vis[N],n, m;

struct node
{
    int x,step;
};

queue<node> q;

bool judge_prime(int x) //判断素数
{
    if(x == 0 || x == 1)
        return false;
    else if(x == 2 || x == 3)
        return true;
    else
    {
        for(int i = 2; i <= (int)sqrt(x); i++)
            if(x % i == 0)
                return false;
        return true;
    }
}

void BFS()
{
    int X, STEP, i;
    while(!q.empty())
    {
        node tmp;
        tmp = q.front();
        q.pop();
        X = tmp.x;
        STEP = tmp.step;
        if(X == m)
        {
            printf("%d\n",STEP);
            return ;
        }
        for(i = 1; i <= 9; i += 2) //个位
        {
            int s = X / 10 * 10 + i;
            if(s != X && !vis[s] && judge_prime(s))
            {
                vis[s] = 1;
                node temp;
                temp.x = s;
                temp.step = STEP + 1;
                q.push(temp);
            }
        }
        for(i = 0; i <= 9; i++) //十位
        {
            int s = X / 100 * 100 + i * 10 + X % 10;
            if(s != X && !vis[s] && judge_prime(s))
            {
                vis[s] = 1;
                node temp;
                temp.x = s;
                temp.step = STEP + 1;
                q.push(temp);
            }
        }
        for(i = 0; i <= 9; i++) //百位
        {
            int s = X / 1000 * 1000 + i * 100 + X % 100;
            if(s != X && !vis[s] && judge_prime(s))
            {
                vis[s] = 1;
                node temp;
                temp.x = s;
                temp.step = STEP + 1;
                q.push(temp);
            }
        }
        for(i = 1; i <= 9; i++) //千位
        {
            int s = i * 1000 + X % 1000;
            if(s != X && !vis[s] && judge_prime(s))
            {
                vis[s] = 1;
                node temp;
                temp.x = s;
                temp.step = STEP + 1;
                q.push(temp);
            }
        }
    }
    printf("Impossible\n");
    return ;
}

int main()
{
    int t, i;
    scanf("%d",&t);
    while(t--)
    {
        while(!q.empty()){
           q.pop();
        }
        scanf("%d%d",&n,&m);
        memset(vis,0,sizeof(vis));
        vis[n] = 1;
        node tmp;
        tmp.x = n;
        tmp.step = 0;
        q.push(tmp);
        BFS();
    }
    return 0;
}