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;
原文：https://www.cnblogs.com/zhgyki/p/9505027.html