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);
		else
			bfs();
	}
	return 0;
}

//后来写的：

#include <cstdio>
#include <cstring>
#include <cmath>
#include <iostream>
#include <queue>
using namespace std;
int prime[10000],book[10000];
struct node{
	int num,step;
	node(int a,int b):num(a),step(b){}
};
void dfs(int s,int t){
	int te,k,dig[4];
	queue<node> q;
	q.push(node(s,0));
	memset(book,0,sizeof(book));
	book[s] = 1;
	while(!q.empty()){
		node h = q.front();
		q.pop();
		te = h.num;
		if(te == t){
			printf("%d\n",h.step);
			return;
		}
		k = 0;
		while(te){
			dig[k++] = te % 10;
			te /= 10;
		}
		for(int i = 0;i < 4;i++){
			for(int j = 0;j <= 9;j++){
				if(j != dig[i] && (i != 3 || j != 0)){
					int a = 0;
					for(int t = 3;t > i;t--){
						a = a*10 + dig[t];
					}
					a = a*10 + j;
					for(int t = i - 1;t >= 0;t--){
						a = a*10 + dig[t];
					}
					if(a >=1000 && !book[a] && !prime[a]){
						book[a] = 1;
						q.push(node(a,h.step + 1));
					}
				}
			}
		}
	}
}
int main(){
	int k,s,t;
	memset(prime,0,sizeof(prime));
	prime[1] = 1;
	for(int i = 2;i < sqrt(10000);i++){
		if(!prime[i]){
			for(int j = i + i;j < 10000;j += i){
				prime[j] = 1;
			}
		}
	}
	scanf("%d",&k);
	for(int i = 1;i <= k;i++){
		scanf("%d%d",&s,&t);
		dfs(s,t);
	}
	return 0;
}