poj3126 Prime Path

题意：

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

思路：

bfs

实现：

#include <iostream>
#include <cstdio>
#include <queue>
#include <cstring>
using namespace std;

const int MAXN = 10000;
bool check[MAXN + 5];
bool vis[MAXN + 5];

struct node
{
    int now, d;
};

void init()
{
    for (int i = 0; i <= MAXN; i++)
    {
        check[i] = true;
    }
    check[0] = check[1] = false;
    for (int i = 2; i <= MAXN; i++)
    {
        if (check[i])
        {
            for (int j = 2 * i; j <= MAXN; j += i)
                check[j] = false;
        }
    }
}

bool is_prime(int x)
{
    return check[x];
}

int solve(int x, int t)
{
    node start;
    start.now = x;
    start.d = 0;
    vis[x] = true;
    queue<node> q;
    q.push(start);
    while (!q.empty())
    {
        node tmp = q.front();
        q.pop();
        if (tmp.now == t)
        {
            return tmp.d;
        }
        int u = tmp.now;
        int t = 1000;
        int last = 0;
        for (int i = 0; i < 4; i++)
        {
            int now = u / t % 10;
            for (int j = 0; j <= 9; j++)
            {
                if ((i == 0 && j == 0) || j == now)
                    continue;
                int n = last + j * t + u % t;
                if (is_prime(n) && !vis[n])
                {
                    vis[n] = true;
                    node son;
                    son.now = n;
                    son.d = tmp.d + 1;
                    q.push(son);
                }
            }
            last += now * t;
            t /= 10;
        }
    }
    return -1;
}

int main()
{
    int t, x, y;
    cin >> t;
    init();
    while (t--)
    {
        cin >> x >> y;
        memset(vis, 0, sizeof(vis));
        int res = solve(x, y);
        if (res != -1)
            cout << res << endl;
        else
            cout << "Impossible" << endl;
    }
    return 0;
}