POJ3126——BFS——Prime Path

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

Source

Northwestern Europe 2006

大意：从一个素数到另一个素数，每次只能改变一个数，要求每次改变后的数都是素数，要求算出转换的最小次数，简单BFS，不过注意d要还原

#include<cstdio>

#include<cstring>

#include<algorithm>

#include<queue>

using namespace std;

const int MAX = 10000;

bool prime[MAX];

void pri()

{

    int i,j;

    for( i = 1000; i < MAX; i++){

       for( j = 2; j < i;j++){

            if(i%j == 0){

                prime[i] = false;

              break;}

       }

       if( i == j)

        prime[i] = true;

    }

}

int d[5];

int num[MAX],k[MAX];

int bfs(int first,int last)

{

    int vis;

    memset(num,0,sizeof(num));

    memset(k,-1,sizeof(k));

    queue <int> q;

    while(!q.empty())

        q.pop();

    q.push(first);

    k[first] = 0;

    while(!q.empty()){

       int now = q.front();

        q.pop();

        d[0] = now/1000;

        d[1] = now%1000/100;

        d[2] = now%100/10;

        d[3] = now%10;

        for(int i = 0; i <4; i++){

            int temp = d[i];

            for(int j = 0; j <= 9;j++){

                 if(d[i] == j) continue;

                d[i] = j;

                 vis = d[0]*1000+d[1]*100+d[2]*10+d[3];

                if(prime[vis] == true && k[vis] == -1){

                    q.push(vis);

                    k[vis] = 0;

                    num[vis] = num[now] + 1;

            }

            if(vis == last) return num[vis];

          }

        d[i] = temp;

        }

    if(now == last) return num[now];

    }

    return -1;

}

int main()

{

    int T,n,m;

    scanf("%d",&T);

    pri();

    while(T--){

        scanf("%d%d",&n,&m);

        if(bfs(n,m)== -1) printf("Impossible\n");

        else

       printf("%d\n",bfs(n,m));

    }

    return 0;

}

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