poj3126(Prime Path)广搜+素数判定

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

大水题，40入口的BFS，剪枝后远远没有40入口。

#include <iostream>
#include <sstream>
#include <fstream>
#include <string>
#include <map>
#include <vector>
#include <list>
#include <set>
#include <stack>
#include <queue>
#include <deque>
#include <algorithm>
#include <functional>
#include <iomanip>
#include <limits>
#include <new>
#include <utility>
#include <iterator>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cmath>
#include <ctime>
using namespace std;

const int maxp = 10010;

bool isPrime[maxp];
int a, b;

void init()
{
    fill(isPrime, isPrime+maxp, true);
    for (int i = 2; i < maxp; ++i)
        if (isPrime[i])
            for (int j = i*i; j < maxp; j += i)
                isPrime[j] = false;
}

int bfs()
{
    int vis[maxp];
    memset(vis, -1, sizeof(vis));
    queue<int> q;
    q.push(a);
    vis[a] = 0;
    while (!q.empty())
    {
        int num = q.front();
        q.pop();
        if (num == b)
            return vis[num];
        for (int i = 0; i < 4; ++i)
            for (int j = 0; j < 10; ++j)
                if (i != 3 || j)
                {
                    int t = (int)(pow(10, i) + 0.5);
                    int c = num - ((num / t) % 10) * t + j * t;
                    if (isPrime[c] && vis[c] == -1)
                    {
                        q.push(c);
                        vis[c] = vis[num] + 1;
                    }
                }
    }
    return -1;
}

int main()
{
    init();
    int T;
    cin >> T;
    while (T--)
    {
        scanf("%d%d", &a, &b);
        int ans = bfs();
        if (ans == -1)
            printf("Impossible
");
        else
            printf("%d
", bfs());
    }
    return 0;
}