POJ 3216 Prime Path （BFS）

Posted 2020-10-01 Farewell

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
题意：
　　每一步只能更改其中的一个数字，并且更改后的数字必须是素数，问最短需要多少步可以变成需要的数字。
题解：
　　求这种最短路一般都用BFS。代码稍微复杂了点。可以打张素数表，可以节省时间。

#include<iostream>
#include<queue>
#include<cstring>
using namespace std;
const int maxn=1e4+5;
bool isp[maxn],vis[maxn];
int n,m;
struct node
{
    int a[4],step;
};
bool is_prime(int n)
{
    for(int i=2;i*i<=n;i++)
    {
        if(n%i==0)
            return false;
    }
    return true;
}
int bfs(int n)
{
    node now;
    now.a[0]=n/1000;
    now.a[1]=n/100%10;
    now.a[2]=n/10%10;
    now.a[3]=n%10;
    now.step=0;
    queue<node>que;
    que.push(now);
    while(que.size())
    {
        now=que.front();
        que.pop();
        node next;
        if(now.a[0]==m/1000&&now.a[1]==m/100%10&&now.a[2]==m/10%10&&now.a[3]==m%10)
            return now.step;
        for(int i=0;i<4;i++)
        {
            for(int j=0;j<10;j++)
            {
                next=now;
                if(i||j)
                    next.a[i]=j;
                else
                    continue;
                int sum=next.a[0];
                for(int k=1;k<4;k++)
                    sum=sum*10+next.a[k];
                if(!vis[sum]&&isp[sum])
                {
                    vis[sum]=true;
                    next.step++;
                    que.push(next);
                }
            }
        }
    }
    return -1;
}
int main()
{
    for(int i=1000;i<10000;i++)
        isp[i]=is_prime(i);//打素数表
    int t;
    cin>>t;
    while(t--)
    {
        memset(vis,false,sizeof(vis));
        cin>>n>>m;
        vis[n]=true;
        cout<<bfs(n)<<endl;
    }
    return 0;
}