Prime Path

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.

这道题就是一个暴力题，我们首先把所有的四位质数存起来（是质数就为1，不是质数就为0），

然后在一位一位的枚举并改动（改出的数是质数我们才改，不然就违背题意了），最后求出答案时就挑出。当然，我说的暴力不是真正的暴力，而是要一位一位地枚举。当然，因为这道题是一个最短路径问题，所以我们也应该用宽搜解决这道题。

#include<iostream>
#include<queue>
#include<cmath>
using namespace std;
queue<int>sm;
int a[10000]={0},d[10000]={0},s,t,n;
int search(int f1,int f2){
	for(int i=1000;i<=9999;i++)//初始化
	{
		d[i]=123456789;
	}
	d[f1]=0;
	sm.push(f);
	while(sm.size()){
		int p=sm.front();sm.pop();
		if(p==f2){
		break;
		}
		int h[5]={0};
		int l=p,wei=0;
		while(l){
			h[++wei]=l%10;
			l/=10;
		} //存下每一位
	    for(int i=0;i<=9;i++){
	    	int ss=i+h[2]*10+h[3]*100+h[4]*1000; //改个位
	    	if(a[ss]==1&&d[ss]>(1+d[p])){
	    	d[ss]=d[p]+1;
			sm.push(ss);	
			}
		}
		
		for(int i=0;i<=9;i++){
	    	int ss=h[1]+i*10+h[3]*100+h[4]*1000; //改十位
	    	if(a[ss]==1&&d[ss]>(1+d[p])){
	    	d[ss]=d[p]+1;
			sm.push(ss);	
			}
		}
		for(int i=0;i<=9;i++){
	    	int ss=h[1]+h[2]*10+i*100+h[4]*1000; //改百位
	    	if(a[ss]==1&&d[ss]>(1+d[p])){
	    	d[ss]=d[p]+1;
			sm.push(ss);	
			}
		}
			for(int i=1;i<=9;i++){//改千位，千位不能为0！
	    	int ss=h[1]+h[2]*10+h[3]*100+i*1000; 
	    	if(a[ss]==1&&d[ss]>(1+d[p])){
	    	d[ss]=d[p]+1;
			sm.push(ss);	
			}
		}
	}
	return d[f2];
}
int main() {
cin>>n;
for(int i=1000;i<=9999;i++){//存下所有四位质数
	int k=sqrt(i);
	for(int j=2;j<=k;j++){
		if(i%j==0){
			goto haha;
		}
	}
	a[i]=1;
	haha:;
}
for(int i=1;i<=n;i++){
	cin>>s>>t;
	int ll=search(s,t); 
	if(ll!=123456789)
	cout<<ll<<endl;
	else cout<<"Impossible"<<endl;//其实没有这种情况
	while(sm.size()){//每执行完一次就要清空，不然会影响下一次的结果
		sm.pop();
	}
}
}