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Prime Path

Time Limit: 1000MS	?	Memory Limit: 65536K
Total Submissions: 26158	?	Accepted: 14376

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.
1033
1733
3733
3739
3779
8779
8179The 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

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

Source

Northwestern Europe 2006

題意：

給定兩個素數a b，求a變到b需要幾步，并且變換時只有一個數字不同，并且是素數

思路：

因為是四位的素數，所以先對素數打表，然后BFS，每次進入隊列的是：只有一個數字不同的四位數，并且是素數，就入隊

#include<iostream> #include<cstring> #include<queue> #include<cmath> #include<cstdio> using namespace std; #define MAXV 10009bool prime[MAXV];void init(){ //對素數打表int i,j;for(i=1000;i<=MAXV;i++){for(j=2;j<i;j++)if(i%j==0){prime[i]=false;break;}if(j==i) prime[i]=true;}/*memset(prime,0,sizeof(prime));int m=(int)sqrt(MAXV+0.5);for(int i=2;i<=m;i++)if(!prime[i])for(int j=i*i;j<MAXV;j+=i)prime[j]=1;*/ } int bfs(int first,int last){bool dis[MAXV];queue <int>q;int v,i,j,temp,vtemp,count[MAXV],t[4];memset(dis,false,sizeof(dis));memset(count,0,sizeof(count));q.push(first);dis[first]=true;while(!q.empty()){v=q.front();q.pop();t[0]=v/1000;t[1]=v%1000/100;t[2]=v%100/10;t[3]=v%10; // printf("%d %d %d %d",t[0],t[1],t[2],t[3]);for(j=0;j<4;j++){temp=t[j];for(i=0;i<10;i++)if(i!=temp){t[j]=i;vtemp=t[0]*1000+t[1]*100+t[2]*10+t[3];if(!dis[vtemp] && prime[vtemp]){count[vtemp]=count[v]+1;dis[vtemp]=true;q.push(vtemp);}if(vtemp==last) return count[vtemp];}t[j]=temp;}if(v==last) return count[v];}return -1; }int main(){int n,a,b,key;init();scanf("%d",&n);while(n--){scanf("%d%d",&a,&b);key=bfs(a,b);if(key!=-1) printf("%d\n",key);else printf("Impossible\n");}return 0; }

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