栈

关于素数打表和最短路径（宽度优先搜索）
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

这道题目的意思是给定两个数，求它们之间的最短路径，但是要求路径全部为素数并且每次只能够变换一位数字（狗比的要求）。解决问题的办法是首先要用一个数组来存储所有1000到9999（题目说明为4位数）然后用bfs来寻找这一最短路径。
源代码：

#include <iostream>  
#include <algorithm>  
#include <cstring>  
#include <cstdio>  
#include <math.h>  
#include <cstdlib>  
#include <queue>  
using namespace std;  

#define N 10100  
int prime[N];  

struct node  ///BFS记录步数，习惯用结构体，比较清晰，话说用数组我还不一定写的出来啊....  
{  
    int x,step;  
};  

void init() ///打N以内的素数表，prime[i]=0为素数。  
{  
    int i,j;  
    memset(prime,0,sizeof(prime));  
    prime[1]=1;  
    for(i=2; i<N; i++)  
    {  
        if(prime[i]==0)  
        {  
            for(j=2; j*i<N; j++)  ///傻逼XX的这里的J是从２开始的。这么简单的地方也会错，唉．．．．  
                prime[j*i]=1;  
        }  
    }  
}  

int BFS(int s,int e)  
{  
    int vis[N],num;　　///vis[N]用来标记是否查找过  
    memset(vis,0,sizeof(vis));  
    queue<node>Q;  ///申请队列。  
    node p,q;    
    p.x=s;     ///将起始值赋值给P  
    p.step=0;  
    vis[s]=1;  
    Q.push(p);  ///将p压入队列  
    while(!Q.empty())  ///当队列不为空时  
    {  
        p=Q.front();  ///取队首  
        Q.pop();    ///清队首  
        if(p.x==e)    ///若找到终止值 返回步数，  
            return p.step;   

        int t[5];  
        t[1]=p.x/1000;   ///记录千位数  
        t[2]=p.x/100%10;  ///记录百位数  
        t[3]=p.x/10%10;   ///记录十位  
        t[4]=p.x%10;   ///记录各位  

        for(int i=1;i<=4;i++)  
        {  
            int temp=t[i];    ///这里特别注意，每次只能变换一位数，这里用来保存该变换位的数，变换下一位数时，该位数要恢复  
            for(int j=0;j<10;j++)  
            {  
                if(t[i]!=j)  
                {  
                    t[i]=j;  ///t[i]要改变  
                    num=t[1]*1000+t[2]*100+t[3]*10+t[4];  
                }  
                if(num>=1000 && num<=9999 && !vis[num] && !prime[num])   ///判断该4位数是否满足条件，  
                {  
                    q.x=num;  
                    q.step=p.step+1;  ///若满足，步数加1，压入队列，标记...  
                    Q.push(q);  
                    vis[num]=1;  
                }  
            }  
            t[i]=temp;  ///恢复  
        }  
    }  
    return -1;  
}  

int main()  
{  
    int n,a,b,ans;  
    init();  
    scanf("%d",&n);  
    while(n--)  
    {  
        scanf("%d%d",&a,&b);  
        ans=BFS(a,b);  
        if(ans!=-1)  
            printf("%d\n",ans);  
        else  
            printf("Impossible\n");  
    }  
}