hdu 1973 Prime Path

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

Sample Output

6
7
0

这题是求从第一个数字变为第二个数字最少需要变几次，即最短路，每一次只能改变一个数字，必须是从素数到素数的改变，bfs遍历，找到最小的步数。所有的都是四位数字，即1000-9999。


代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
struct que {
    int d,times;///d是数字 times是变换次数
}temp;
int vis[9000];
int main() {
    int prime[10000] = {1,1};
    for(int i = 2;i <= 9999;i ++) {///筛选素数表 prime为0表示素数
        if(!prime[i]) {
            for(int j = i;j * i <= 9999;j ++)
                prime[i * j] = 1;
        }
    }
    int T,a,b;
    scanf("%d",&T);
    while(T --) {
        scanf("%d%d",&a,&b);///从a到b的转变
        int flag = 0,d,times,num;
        memset(vis,0,sizeof(vis));///标记数组归0
        queue<que> q;
        temp.d = a;
        temp.times = 0;
        q.push(temp);
        vis[a - 1000] = 1;///节约空间
        while(!q.empty()) {
            if(q.front().d == b) {
                flag = 1;
                printf("%d
",q.front().times);
                break;
            }
            d = q.front().d;
            times = q.front().times;
            for(int i = 1;i <= 1000;i *= 10) {///四位选择一位
                num = d - d / i * i % (i * 10);
                for(int j = 0;j < 10;j ++) {
                    int tnum = num + i * j;
                    if(tnum < 1000 || prime[tnum] || vis[tnum-1000])continue;
                    vis[tnum - 1000] = 1;
                    temp.d = tnum;
                    temp.times = times + 1;
                    q.push(temp);
                }
            }
            q.pop();
        }
        if(!flag) printf("impossible
");
    }
}