Prime Path
F - 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.
— 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.
1033The 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.
1733
3733
3739
3779
8779
8179
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：
给两个素数 num1 num2 要求将num1变成num2 ，每次变换只能变一个数字，花费为一，变换后的数字依旧为素数，求最小花费。
思路：欧拉筛之后，bfs 分别跑个十百千位上的数字，个位数字一定为奇数，千位数字必定不为零。
#include <cstdio> #include <cstring> #include <algorithm> #include <functional> #include <iostream> #include <queue> using namespace std; const int maxn=1e6+10; int prime[maxn],pn; bool noot[maxn]; void init(){ pn=0; memset(noot,false,sizeof(noot)); noot[1]=1; for (int i=2;i<maxn;i++){ if(!noot[i])prime[pn++]=i; for (int j=0;j<pn&&i*prime[j]<maxn;j++){ noot[i*prime[j]]=1; if(i%prime[j]==0) break; } } } struct node{ int num,step; }now,nxt; int num1,num2; int bfs(int x){ bool vis[maxn]; memset(vis,false,sizeof(vis)); vis[x]=1; now.num=x; now.step=0; queue<node>q; q.push(now); while(!q.empty()){ now=q.front(); q.pop(); if(now.num==num2) return now.step; for (int i=1;i<=9;i+=2){ int num=now.num/10*10+i; if(!noot[num]&&!vis[num]){ nxt.num=num; nxt.step=now.step+1; vis[num]=1; q.push(nxt); } }//个位 for (int i=0;i<=9;i++){ int num=now.num%10+(now.num/100*10+i)*10; if(!noot[num]&&!vis[num]){ nxt.num=num; nxt.step=now.step+1; vis[num]=1; q.push(nxt); } }//十位； for (int i=0;i<=9;i++){ int num=now.num%100+(now.num/1000*10+i)*100; if(!noot[num]&&!vis[num]){ nxt.num=num; nxt.step=now.step+1; vis[num]=1; q.push(nxt); } } for (int i=1;i<=9;i++){ int num=now.num%1000+i*1000; if(!noot[num]&&!vis[num]){ nxt.num=num; nxt.step=now.step+1; vis[num]=1; q.push(nxt); } } } return -1; } int main(){ init(); int t; scanf("%d",&t); while(t--){ scanf("%d%d",&num1,&num2); int ans=bfs(num1); if(ans>=0) printf("%d\n",ans); else printf("Impossible\n"); } return 0; }
