Prime Path--POJ3126(bfs)

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

题意：给你两个四位数m和n，求m变到n至少需要几步；每次只能从个十百千上改变一位数字，并且改变后的数要是素数；
我们可以用优先对列做；

#include<iostream>

#include<stdio.h>

#include<string.h>

#include<queue>

#include<cmath>

#define N 11000

using namespace std;

int prime(int n)

{

    int i,k;

    k=(int)sqrt(n);

    for(i=2;i<=k;i++)

        if(n%i==0)

            return 0;

    return 1;

}

struct node

{

    int x,step;

    friend bool operator<(node a,node b)

    {

        return a.step>b.step;

    }

};

int bfs(int m,int n)

{

    int a[4]={1,10,100,1000};

    int vis[N];

    priority_queue<node>Q;

    node q,p;

    memset(vis,0,sizeof(vis));

    vis[m]=1;

    q.x=m;

    q.step=0;

    Q.push(q);

    while(!Q.empty())

    {

        q=Q.top();

        Q.pop();

        if(q.x==n)

            return q.step;

        for(int i=0;i<4;i++)//控制改变的哪一位

        {

            for(int j=0;j<10;j++)//把相对应的那一位变成j；

            {

                int L=q.x/(a[i]*10);

                int R=q.x%(a[i]);

                p.x=L*(a[i]*10)+j*a[i]+R;//重新组成的数；

                if(p.x>=1000&&vis[p.x]==0&&prime(p.x)==1)

                {

                    vis[p.x]=1;

                    p.step=q.step+1;

                    Q.push(p);

                }

            }

        }

    }

    return -1;

}



int main()

{

    int T,m,n,ans;

    scanf("%d",&T);

    while(T--)

    {

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

        ans=bfs(m,n);

        if(ans==-1)

            printf("Impossible\n");

        else

            printf("%d\n",ans);

    }

    return 0;

}

View Code