Poj3126 prime

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 
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
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

#include <stdio.h>#include <string.h>struct node{int number;int step;};struct node queue[10000];int start,end;int book[10000];int prime[10000];int digit[4];int head,tail;void bfs(){int num;head=tail=0;queue[tail].number=start;queue[tail].step=0;book[start]=1;tail++;while(head<tail){digit[0]=queue[head].number/1000;digit[1]=queue[head].number%1000/100;digit[2]=queue[head].number%100/10;digit[3]=queue[head].number%10;for(int i=0;i<4;i++){for(int j=0;j<10;j++){//开始因为这里调试好久，每次变化了digit，要记得恢复int temp=digit[i];digit[i]=j;num=digit[0]*1000+digit[1]*100+digit[2]*10+digit[3];digit[i]=temp;if(num>=1000&&num!=queue[head].number&&!book[num]&&!prime[num]){book[num]=1;queue[tail].number=num;queue[tail].step=queue[head].step+1;if(queue[tail].number==end){printf("%d\n",queue[tail].step);return;}tail++;}}}head++;}printf("Impossible\n");return;}int main(){int n;scanf("%d",&n);memset(prime,0,sizeof(prime));prime[0]=prime[1]=1;for(int i=2;i<=5000;i++){if(!prime[i]){for(int j=i+i;j<10000;j+=i)prime[j]=1;}}for(int i=1;i<=n;i++){memset(book,0,sizeof(book));scanf("%d%d",&start,&end);if(start==end)printf("%d\n",0);elsebfs();}return 0;}