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.

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

31033 81791373 80171033 1033

Sample Output

670

题意  给 你两个数n，m，变化n中的一个数字（千位不能为0）并且每次变化都为质数，求最短的变化次数，，

思路  类似于求迷宫最短逃生路径，应用bfs遍历每一种情况 求最少变化次数，

           先用素数打掉，判断此数字是否为质数，再用一个数字用于标记此数字是否进入队列进入则不会再进，用num【】数组记录次数、

注意  对于个位来说不能为偶数 ，若个位是偶数他一定不会是质数，对于千位不能为0，  十位与百位没有限制

#include<cstdio>#include<cstring>#include<queue>using namespace std;const int M=10000;int su[M];int  vis[M];int num[M];void init(){    memset(su,0,sizeof(su));    for(int i=2;i*i<M;i++)        for(int j=2*i;j<M;j+=i)        su[j]=1;        return;}int main(){    int n,m,f;    int t;    init();    queue<int>que;    scanf("%d",&t);    while(t--)    {        f=0;        scanf("%d%d",&n,&m);        memset(vis,0,sizeof(vis));        memset(num,0,sizeof(num));        while(!que.empty())            que.pop();        que.push(n);        vis[n]=1;        while(!que.empty())        {            int tmp=que.front();            que.pop();            if(tmp==m)            {                printf("%d\n",num[m]);                f=1;                break;            }            //个位            for(int i=1;i<=9;i+=2)            {                int res=tmp/10*10+i;                if(vis[res]==0&&su[res]==0)                {                    vis[res]=1;                    que.push(res);                    num[res]=num[tmp]+1;                }            }            //十位            for(int i=0;i<=9;i++)            {                int res=tmp/100*100+i*10+tmp%10;                if(vis[res]==0&&su[res]==0)                {                    vis[res]=1;                    que.push(res);                    num[res]=num[tmp]+1;                }            }            //百位            for(int i=0;i<=9;i++)            {                int res=tmp/1000*1000+i*100+tmp%100;                if(vis[res]==0&&su[res]==0)                {                    vis[res]=1;                    que.push(res);                    num[res]=num[tmp]+1;                }            }            //千位            for(int i=1;i<=9;i++)            {                int res=i*1000+tmp%1000;                if(vis[res]==0&&su[res]==0)                {                    vis[res]=1;                    que.push(res);                    num[res]=num[tmp]+1;                }            }        }        if(f==0)            printf("Impossible\n");    }    return 0;}