POJ 3126：Prime Path（素数+BFS）

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.

Examples

Input

3
1033 8179
1373 8017
1033 1033

Output

6
7
0

题意

给出两个四位的素数$n,m$，要求$n$每次只能变换一位，并且变换后的数字依旧是素数。
求$n$经过多少步变换能够变成$m$；如果$n$无法变成$m$，输出Impossible

思路

将$n$的四位数字拆分了，每次变换一位，来判断是否符合条件，如果符合条件，将新数字加入队列，至到数字和$m$相等，或队列为空

AC代码

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <math.h>
#include <limits.h>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#include <set>
#include <string>
#include <time.h>
#define ll long long
#define ull unsigned long long
#define ms(a,b) memset(a,b,sizeof(a))
#define pi acos(-1.0)
#define INF 0x7f7f7f7f
#define lson o<<1
#define rson o<<1|1
#define bug cout<<"-------------"<<endl
#define debug(...) cerr<<"["<<#__VA_ARGS__":"<<(__VA_ARGS__)<<"]"<<"\n"
const int maxn=1e4+10;
const int mod=1e9+7;
using namespace std;
int vis[maxn];
int _vis[maxn];
int cnt;
struct node
{
	int num;
	int step;
};
int bfs(int n,int m)
{
	ms(_vis,0);
	int newnum;
	node p,q;
	queue<node>que;
	p.num=n;
	p.step=0;
	que.push(p);
	_vis[n]=1;
	while(!que.empty())
	{
		q=que.front();
		que.pop();
		if(q.num==m)
			return q.step;
		int get[4];
		int N=q.num;
		int _=0;
		while(N)
		{
			get[_++]=N%10;
			N/=10;
		}
		for(int i=0;i<4;i++)
		{
			int __=get[i];
			for(int j=0;j<=9;j++)
			{
				if(get[i]!=j)
				{
					get[i]=j;
					newnum=get[0]+get[1]*10+get[2]*100+get[3]*1000;
				}
				if(!vis[newnum]&&newnum>=1000&&newnum<10000&&!_vis[newnum])
				{
					p.num=newnum;
					p.step=q.step+1;
					_vis[newnum]=1;
					que.push(p);
				}
			}
			get[i]=__;
		}
	}
	return -1;
}
void init()
{
	vis[0]=vis[1]=1;
	for(int i=2;i<maxn;i++)
	{
		if(!vis[i])
		{
			for(int j=2;j*i<maxn;j++)
				vis[i*j]=1;
		}
	}
}
int main(int argc, char const *argv[])
{
	ios::sync_with_stdio(false);
	#ifndef ONLINE_JUDGE
	    freopen("in.txt", "r", stdin);
	    freopen("out.txt", "w", stdout);
	    double _begin_time = clock();
	#endif
	init();
	int t;
	cin>>t;
	while(t--)
	{
		int n,m;
		cin>>n>>m;
		int ans=bfs(n,m);
		if(ans==-1)
			cout<<"Impossible"<<endl;
		else
			cout<<ans<<endl;
	}
	#ifndef ONLINE_JUDGE
	    long _end_time = clock();
	    printf("time = %lf ms.", _end_time - _begin_time);
	#endif
	return 0;
}