POJ3126

A - Prime Path
Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

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

题目大意：给出两个质数a,b。从a变化到b每次只能对一位数字进行操作，且操作后的数字依然是质数，变化到b的最小操作次数。

//poj 3126#include "iostream"#include "cstdio"#include "queue"using namespace std;int ncase;int prime_1,prime_2;struct ehh{    int nownumber,step;}m;bool a[10000];queue<ehh> q;bool is_prime(int n) //判断是否为质数{    int result = 0;    int i;    if (n <= 1)    {        return result;    }    if(2 == n)    {        result = 1;        return result;    }    for (i = 2; i < n ;i++)    {        if (0 == n % i)        {            result = 0;            break;        }        else if (i + 1 == n)        {            result = 1;        }        else        {            continue;        }    }    return result;}int ge,shi,bai,qian,store;bool flag=false;void bfs(){    while(!q.empty())    {        m=q.front();        if(m.nownumber==prime_2)        {            cout<<m.step<<endl;            flag=true;            break;        }        q.pop();        ehh temp;        temp=m;        int n;        for(int j=1;j<=4;j++) //只改变一位数字时的所有可能        {            for(int i=0;i<=9;i++)            {                if(j==1)              {                  n=m.nownumber/10;                  ge=m.nownumber%10;                  n=n*10;                  n=n+i;                  if(a[n]==false)                  {                      a[n]=true;                      if(is_prime(n))                      {                          temp.nownumber=n;                          temp.step=m.step+1;                          q.push(temp);                      }                  }              }                if(j==2)                {                   n=m.nownumber/100;                   ge=m.nownumber%10;                   n=n*100+ge;                   n=n+i*10;                   if(a[n]==false)                      {                          a[n]=true;                          if(is_prime(n))                          {                              temp.nownumber=n;                              temp.step=m.step+1;                              q.push(temp);                          }                      }                }                if(j==3)                {                    n=m.nownumber/1000;                    ge=m.nownumber%10;                    shi=m.nownumber%100/10;                    n=n*1000+ge+shi*10;                    n=n+i*100;                    if(a[n]==false)                    {                        a[n]=true;                        if(is_prime(n))                        {                            temp.nownumber=n;                            temp.step=m.step+1;                            q.push(temp);                        }                    }                }                if(j==4)                {                    ge=m.nownumber%10;                    shi=m.nownumber%100/10;                    bai=m.nownumber%1000/100;                    if(i!=0)  //首位不为0                    {                        n=i*1000+ge+shi*10+bai*100;                        if(a[n]==false)                        {                            a[n]=true;                            if(is_prime(n))                            {                                temp.nownumber=n;                                temp.step=m.step+1;                                q.push(temp);                            }                        }                    }                }            }    }    }}int main(){   while(scanf("%d",&ncase)!=EOF)   {       for(int i=1;i<=ncase;i++)       {           while(!q.empty())               q.pop();           flag=false;           cin>>prime_1>>prime_2;           memset(a,false,sizeof(a));           m.nownumber=prime_1;           m.step=0;           a[prime_1]==true;           q.push(m);           bfs();           if(flag==false)               cout<<"Impossible"<<endl;       }   }}