题目传送门

 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.

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

Output

3
1033 8179
1373 8017
1033 1033

Sample Output

6
7
0题意：给你四位数字a,b;每一不只能改变一位数字，且新的数字只能是素数，要你输出最小步数

题解：bfs，每次向下遍历40个方向

代码：

#include<iostream>
#include<queue>
#include<string.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define mod 1000000007
#define INF 0x3f3f3f3f
struct niu
{int prime,step;niu(){}niu(int pr,int st){prime=pr,step=st;}
};
int a,b;
int ans=INF;
bool check(int a)
{for(int i=2;i*i<=a;i++)if(a%i==0)return false;return a!=1;
}
queue<niu>q;
bool vis[10005];
int bfs()
{memset(vis,false,sizeof(vis));while(q.size())q.pop();q.push(niu(a,0));vis[a]=true;while(q.size()){niu tmp=q.front();q.pop();//cout<<tmp.prime<<tmp.step<<endl;if(tmp.prime==b){ans=min(ans,tmp.step);}int cnt=tmp.prime%10;int cur=(tmp.prime/10)%10;for(int i=0;i<=9;i++){int nx=(tmp.prime/10)*10+i;if(check(nx)&&!vis[nx]){vis[nx]=true;q.push(niu(nx,tmp.step+1));}int ny=(tmp.prime/100)*100+i*10+cnt;if(check(ny)&&!vis[ny]){vis[ny]=true;q.push(niu(ny,tmp.step+1));}int nz=(tmp.prime/1000)*1000+i*100+cur*10+cnt;if(check(nz)&&!vis[nz]){vis[nz]=true;q.push(niu(nz,tmp.step+1));}if(i==0)continue;int nn=(tmp.prime%1000)+i*1000;//cout<<nn<<endl;if(check(nn)&&!vis[nn]){vis[nn]=true;q.push(niu(nn,tmp.step+1));}}}return ans==INF?-1:ans;
}
int main()
{int T;cin>>T;while(T--){ans=INF;//注意这里的ans要初始化cin>>a>>b;if(bfs()==-1)cout<<"Impossible"<<endl;else cout<<bfs()<<endl;}return 0;