poj 3126 Prime Path (bfs~~)

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<cstdio>
#include<cstring>
#include<ctime>
#include<iostream>
#include<algorithm>
#include<string>
#include<queue>
#define LL long long
using namespace std;
int ss[10005] = { 0 };
int cnt = 0;//素数个数
int zs[10000];//保存着四位素数
void sss()//素数晒先筛出四位的素数
{
	for (int s = 2; s < 10002; s++)
	{
		if (ss[s] == 0)
		{
			if (s >= 1000)
			{
				zs[cnt++] = s;
			}
			for (int k = 2 * s; k < 10002; k += s)
			{
				ss[k] = 1;
			}
		}
	}
}
bool vis[10005] = { 0 };//防止多次便利
int maxn = 99999;
int st, en;
bool check(int a, int b)
{
	int spot = 0;
	while (a)
	{
		if (a % 10 != b % 10)
		{
			spot++;
		}
		a = a / 10;
		b = b / 10;
	}
	if (spot == 1)
	{
		return 1;
	}
	return 0;
}
struct ***
{
	int data;//素数的值
	int step;//步数
};
queue<***>q;
void bfs()
{
	while (!q.empty())
	{
		*** t = q.front();
		q.pop();
		if (t.data == en)
		{
			maxn = t.step;
			return;//注意这里~~第一次找到之后就立刻return就好（因为这是bfs~最先找到一定路径最短）
		}
		for (int s = 0; s < cnt; s++)
		{
			if (check(zs[s], t.data) && !vis[zs[s]])
			{
				*** w;
				w.data = zs[s];
				w.step = t.step + 1;
				vis[zs[s]] = 1;
				q.push(w);
			}
		}
	}
}
int main()
{
	int te;
	scanf("%d", &te);
	sss();
	while (te--)
	{
		while (!q.empty())
		{
			q.pop();
		}
		maxn = 9999;
		memset(vis, 0, sizeof(vis));
		scanf("%d%d", &st, &en);
		vis[st] = 1;
		*** w;
		w.step = 0;
		while (1)
		{
			break;
		}		w.data = st;

		q.push(w);
		bfs();
		if (maxn != 99999)
		{
			printf("%d\n", maxn);
		}
		else
		{
			printf("Impossible\n");
		}
	}
	return 0;
}