Description
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.
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：一个四位数，一步可以将某一位上的数字变为任意数字，但要保证结果仍然是四位素数，求将一个四位素数变为另一个四位素数的最小步骤数。
思路分析：赤裸裸的BFS。预先处理出素数。
代码如下：
1 # include<iostream> 2 # include<cstdio> 3 # include<queue> 4 # include<cstring> 5 # include<algorithm> 6 using namespace std; 7 int mark[10005],vis[10005],base[4]; 8 struct node 9 { 10 int num,t; 11 node(int a,int b):num(a),t(b){} 12 }; 13 void init() 14 { 15 memset(mark,0,sizeof(mark)); 16 mark[1]=1; 17 for(int i=2;i<=10000;++i){ 18 if(mark[i]) 19 continue; 20 for(int j=2*i;j<=10000;j+=i) 21 mark[j]=1; 22 } 23 base[0]=1; 24 for(int i=1;i<4;++i) 25 base[i]=base[i-1]*10; 26 } 27 void bfs(int s,int e) 28 { 29 memset(vis,0,sizeof(vis)); 30 queue<node>q; 31 vis[s]=1; 32 q.push(node(s,0)); 33 while(!q.empty()){ 34 node u=q.front(); 35 q.pop(); 36 if(u.num==e){ 37 printf("%d ",u.t); 38 return ; 39 } 40 for(int i=0;i<4;++i){ 41 int now=u.num; 42 int cur=now%(base[i]*10); 43 cur/=base[i]; 44 now=now-cur*base[i]; 45 for(int k=0;k<=9;++k){ 46 int temp=now+base[i]*k; 47 if(temp>1000&&!mark[temp]&&!vis[temp]){ 48 vis[temp]=1; 49 q.push(node(temp,u.t+1)); 50 } 51 } 52 } 53 } 54 } 55 int main() 56 { 57 init(); 58 int a,b,T; 59 scanf("%d",&T); 60 while(T--) 61 { 62 scanf("%d%d",&a,&b); 63 bfs(a,b); 64 } 65 return 0; 66 }