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：给你两个四位数素数x，y。求从x到y最少改变几次。改变规则，每次只能改变一位，改变后的x也必须是素数。
思路：因为是求最小次数，用bfs。依次改变个位，十位，百位，千位，满足条件的入队列，直到改变到y为止。（不要忘记vis数组标记。另外，队列里的元素是结构体，同时记录改变后的x和改变所需步数）
如果用c++交的话，sqrt里面的参数须是double或float
注意个位、十位、百位、千位的改变方式。个位、十位、百位是从0开始改变的，千位是从1开始，不是从当前位向后加。
#include<iostream>
#include<cstring>
#include<queue>
#include<cmath>
using namespace std;
int ans;
int vis[10010];
struct date
{
int k;//记录改变后的x
int st;//记录改变经历步数
};
queue<date>q;
void qingkong()
{
while(!q.empty())
{
q.pop();
}
}
int prime(int m)
{
if(m==0||m==1)
return 0;
else if(m==2||m==3)
return 1;
else
{
int flag=1;
for(int i=2;i<=(int)sqrt(1.0*m);i++)
{
if(m%i==0)
{
flag=0;
break;
}
}
return flag;
}
}
void bfs(int x,int y)
{
int i,s;
date u,v;
u.k=x;
u.st=0;
vis[u.k]=1;
q.push(u);
while(!q.empty())
{
v=q.front();
q.pop();
if(v.k==y)
{
cout<<v.st<<endl;
return ;
}
for(i=1;i<=9;i+=2)//个位
{
s=v.k/10*10+i;
if(s!=v.k&&!vis[s]&&prime(s))
{
vis[s]=1;
u.k=s;
u.st=v.st+1;
q.push(u);
}
}
for(i=0;i<=9;i++)//十位
{
s=v.k/100*100+i*10+v.k%10;
if(s!=v.k&&!vis[s]&&prime(s))
{
vis[s]=1;
u.k=s;
u.st=v.st+1;
q.push(u);
}
}
for(i=0;i<=9;i++)
{
s=v.k/1000*1000+i*100+v.k%100;
if(s!=v.k&&!vis[s]&&prime(s))
{
vis[s]=1;
u.k=s;
u.st=v.st+1;
q.push(u);
}
}
for(i=1;i<=9;i++)
{
s=i*1000+v.k%1000;
if(s!=v.k&&!vis[s]&&prime(s))
{
vis[s]=1;
u.k=s;
u.st=v.st+1;
q.push(u);
}
}
}
cout<<"Impossible"<<endl;
return ;
}
int main()
{
int n,x,y;
cin>>n;
while(n--)
{
memset(vis,0,sizeof(vis));
cin>>x>>y;
qingkong();
bfs(x,y);
}
return 0;
}
来源：https://blog.csdn.net/weixin_44145887/article/details/98725545