poj 3126 Prime Path

Prime Path

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 14864		Accepted: 8379

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

比如从1033到8179，，，枚举每一个位数上（左数第1-4）改变后的每一个值，关键是要用到bfs的思想

由斤及远，，在队列中最先找到的必定是最小的step

#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
using namespace std;
struct Node{
     int step;
     bool prime;
}node[10001];
int x[5];
void  init()
{
    for(int i=2;i<=9999;i++)
        node[i].prime=1;
     for(int j=2;j<=9999;j++)
             if(node[j].prime)
             {
                 for(int i=j*2;i<=9999;i+=j)
                     node[i].prime=0;
             }
}
int bfs(int a,int b)
{
    queue<int> q;
    q.push(a);
    int p;
    while(q.size())
    {
        p=q.front();
        q.pop();
        x[4]=p%10;
        x[3]=(p/10)%10;
        x[2]=(p/100)%10;
        x[1]=(p/1000)%10;//取个位，十位，百位，千位
        for(int i=1;i<=4;i++)
        {
           int k=x[i];
           for(int j=0;j<=9;j++)
        {
            if(i==1&&j==0)
                continue;
            x[i]=j;
            int temp=x[1]*1000+x[2]*100+x[3]*10+x[4];
            if(node[temp].prime&&node[temp].step==0&&(temp!=p))//注意排除掉本身的值，否则                                                               会多加个一
             {
               node[temp].step=node[p].step+1;
               if(temp==b)
                 {
                     printf("%d
",node[temp].step);
                     return 1;
                 }
                 q.push(temp);
             }
        }
            x[i]=k;//记得最后要将x[i]的值复原，，不能干扰到下一个i的判断。
        }
    }
    printf("Impossible
");
    return 0;
}
void chu()
{
    for(int i=1000;i<=9999;i++)
             node[i].step=0;
}
int main()
{
     int a,b,cas;
     init();
     cin>>cas;
     while(cas--)
     {
         chu();
         cin>>a>>b;
         if(a==b)
           printf("0
");
         else
         bfs(a,b);
     }
     return 0;
}