标签：who oom put cout The security ted last tin
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;
标签：who oom put cout The security ted last tin
原文地址：https://www.cnblogs.com/zhgyki/p/9505027.html
