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 一看这道题就想到bfs 代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <map>
#include <cmath>
#include <string>
#include <queue>
using namespace std;
class que
{
public:
int d;
int times;
}temp;
int vis[9000],sta,time,num;
int main()
{
int prime[10000]={1,1,0,0};
for(int i=2;i<=9999;i++)
{
if(!prime[i])
{
for(int j=i;j*i<=9999;j++)
prime[i*j]=1;
}
}
int T,a,b;
queue<que>q;
cin>>T;
while(T--)
{
cin>>a>>b;
int flag=0;
memset(vis,0,sizeof(vis));
while(!q.empty())q.pop();
temp.d=a,temp.times=0;
q.push(temp);
vis[a-1000]=1;
if(!prime[a]&&!prime[b])
while(!q.empty())
{
if(q.front().d==b)
{
flag=1;
cout<<q.front().times<<endl;
break;
}
sta=q.front().d,time=q.front().times;
for(int i=1000;i>0;i/=10)
{
for(int j=0;j<10;j++)
{
num=sta-sta/i%10*i+i*j;
if(num<1000||prime[num]||vis[num-1000])continue;
vis[num-1000]=1;
temp.d=num,temp.times=time+1;
q.push(temp);
}
}
q.pop();
}
if(!flag)cout<<"impossible"<<endl;
}
}