POJ-3126-Prime Path

Prime Path

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 16991		Accepted: 9561

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

给定两个四位素数a  b，把a转换成b转换的过程要求每次转换出来的数都是一个 四位素数，而且当前这步的变换所得的素数与前一步得到的素数  只能有一个位不同，而且每步得到的素数都不能重复。

输出从a到b最少需要的变换次数。无法变换则输出Impossible

#include<iostream>
#include<string.h>
#include<stdio.h>
#include<math.h>
#include<algorithm>
#include<queue>
#include<vector>
using namespace std;
bool JudgePrim(int n)
{
    if(n<2)return false;
    else
    {
        for(int i = 2; i<=sqrt(n); ++i)
            if(n%i==0)return false;
        return true;
    }
}
struct node
{
    int data, s;
} p, w;
int f(int x, int i)
{
    int k;
    if(i==0)   k = x-(x%10);
    else if(i==1)    k = x-x%100+x%10;
    else if(i==2)  k =x-x%1000+x%100;
    else   k = x%1000;
    return k;
}
int main()
{
    int n, a, b, k;
    bool vis[10000];
    scanf("%d", &n);
    while(n--)
    {
        scanf("%d%d", &a, &b);
        memset(vis, 0,sizeof(vis));
        vis[a] = 1;
        queue<node>qq;
        p.data = a;
        p.s = 0;
        qq.push(p);
        while(!qq.empty())
        {
            p = qq.front();
            if(p.data==b)
            {
                printf("%d\n", p.s);
                break;
            }
            qq.pop();
            for(int ii = 0; ii<4; ++ii)
            {
                for(int i = 0; i<=9; ++i)
                {
                    if(i==0&&ii==3)continue;
                    k = f(p.data, ii)+pow(10,ii)*i;
                    if(!vis[k]&&JudgePrim(k))
                    {
                        w.data = k;
                        w.s = p.s+1;
                        vis[w.data]  = 1;
                        qq.push(w);
                    }
                }
            }
        }
        if(qq.empty())printf("Impossible\n");
    }
    return 0;
}