POJ 3126 Prime Path( 广搜 )

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.

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).

One line for each case, either with a number stating the minimal cost or containing the word Impossible.

3
1033 8179
1373 8017
1033 1033

6
7
0

#include <stdio.h>
#include <queue>
#include <string.h>
using namespace std;
#define HUR 100
#define THO 1000
#define TEN 10
const int maxn=10000;
bool prime[maxn+5];
int vis[maxn],s,e;

void prime_table(){
	int i,j;
	memset(prime,0,sizeof(prime));
	for(i=2;i<maxn;i++)
		if(!prime[i])
			for(j=i*i;j<maxn;j+=i)
				prime[j]=1;
}

int bfs(){
	int i;
	memset(vis,0,sizeof(vis));
	vis[s]=1;
	queue<int> que;
	que.push(s);
	while(!que.empty()){
		int t=que.front();
		que.pop();
		int d=t;
		d%=1000;
		for(i=1;i<10;i++){
			int tt=d+i*THO;		//变换千位
			if(prime[tt]==0 && vis[tt]==0){
				if(tt==e) return vis[t];
				que.push(tt);vis[tt]=vis[t]+1;
			}
		}
		d=t%100+(t/1000*1000);
		for(i=0;i<10;i++){
			int tt=d+i*HUR;		//变换百位
			if(prime[tt]==0 && vis[tt]==0){
				if(tt==e) return vis[t];
				que.push(tt);vis[tt]=vis[t]+1;
			}
		}
		d=t%10+t/100*100;
		for(i=0;i<10;i++){
			int tt=d+i*TEN;		//变换十位
			if(prime[tt]==0 && vis[tt]==0){
				if(tt==e) return vis[t];
				que.push(tt);vis[tt]=vis[t]+1;
			}
		}
		d=t/10*10;
		for(i=0;i<10;i++){
			int tt=d+i;			//变换个位
			if(prime[tt]==0 && vis[tt]==0){
				if(tt==e) return vis[t];
				que.push(tt);vis[tt]=vis[t]+1;
			}
		}
	}
	return 0;
}

int main()
{
	int T,res;
	prime_table();
	scanf("%d",&T);
	while(T--){
		scanf("%d%d",&s,&e);
		if(s==e){
			printf("0\n");
			continue;
		}
		res=bfs();
		if(res==0)
			printf("Impossible\n");
		else
			printf("%d\n",res);
	}
	return 0;
}