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
题目大意：多组测试数据。给你两个素数n，m，要求每次只改变n的一个数字，且改变后的数仍然是素数，问经过多少次改变后素数n能变成素数m，输出变化最少的次数
思路：因为是求最小次数，用bfs可以保证。所以分别枚举个十百千位的数字+素数判定。
#include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <queue> using namespace std; int t,n,m; const int N=10010; int vis[N]; struct node { int x,step; }; queue<node> q; int pdss(int n) //素数判定 { if(n==0||n==1) return 0; else if(n==2||n==3) return 1; for(int i=2;i*i<=n;++i) { if(n%i==0) return 0; } return 1; } void bfs() { int X,STEP,i,k; node ss,sss; while(!q.empty()) { ss=q.front(); q.pop(); X=ss.x; STEP=ss.step; if(X==m) { printf("%d\n",STEP); return; } for(i=1;i<=9;i+=2) { k=X/10*10+i; if(pdss(k)&&!vis[k]&&k!=X) { vis[k]=1; sss.x=k; sss.step=STEP+1; q.push(sss); } } for(i=0;i<=9;i++) { k=X/100*100+i*10+X%10; if(pdss(k)&&!vis[k]&&k!=X) { vis[k]=1; sss.x=k; sss.step=STEP+1; q.push(sss); } } for(i=0;i<=9;i++) { k=X/1000*1000+i*100+X%100; if(pdss(k)&&!vis[k]&&k!=X) { vis[k]=1; sss.x=k; sss.step=STEP+1; q.push(sss); } } for(i=1;i<=9;i++) { k=i*1000+X%1000; if(pdss(k)&&!vis[k]&&k!=X) { vis[k]=1; sss.x=k; sss.step=STEP+1; q.push(sss); } } } printf("Impossible\n"); return ; } int main(int argc, char *argv[]) { scanf("%d",&t); while(t--) { while(!q.empty()) q.pop(); scanf("%d%d",&n,&m); memset(vis,0,sizeof(vis)); node s; s.x=n; s.step=0; vis[n]=1; q.push(s); bfs(); } return 0; }
