# POJ 3126 Prime Path （BFS）

Prime Path

 Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 18800 Accepted: 10567

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

### AC 代码

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<iostream>
using namespace std;
#include<vector>
#include<queue>

#define Max 10000
int prime[Max];
bool isvis[Max];    //已经访问过的数字不需要重新加入队列
void IsPrime()      //素数打表
{
prime[0]=prime[1]=0;
prime[2]=1;
for(int i=3; i<Max; i++)
prime[i]=i%2==0?0:1;
int t=(int)sqrt(Max*1.0);
for(int i=3; i<=t; i++)
if(prime[i])
for(int j=i*i; j<Max; j+=2*i) //优化
prime[j]=0;
}
struct point
{
int data;
int time;   //当前进行到第几步
point(int data,int time)
{
this->data=data;
this->time=time;
}
};
int solve(point a,int b)
{
memset(isvis,false,sizeof(isvis));
queue<point>sk;
sk.push(a);
while(!sk.empty())
{
point p=sk.front();
sk.pop();
if(p.data==b)return p.time;
isvis[p.data]=true;
for(int i=10; i<=10000; i*=10)  //从低到高枚举
{
for(int j=(i==10000)?1:0; j<=9; j++)    //最高位不能是0
{
int num=p.data/i*i+i/10*j+p.data%(i/10);    //计算当前的数字
if(!isvis[num]&&prime[num])
sk.push(point(num,p.time+1));
}
}
}
return -1;
}

int main()
{
IsPrime();
int n;
while(~scanf("%d",&n))
{
int a,b;
for(int i=0; i<n; i++)
{
scanf("%d%d",&a,&b);
int ans=solve(point(a,0),b);
printf(ans==-1?"Impossible\n":"%d\n",ans);
}
}
return 0;
}