Prime Path bfs搞定

Posted 2020-09-30 给杰瑞一块奶酪~

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;
    }
}