POJ 3126 Prime Path（bfs+素数打表）

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.

1033173337333739377987798179

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

题目大意：给出两个四位数a,b，且都是素数。要求从a变化到b最少需要多少步，每一步可以改动a的任意位上面的数字，但必须保证改变之后的a仍然要为一个素数。

解题思路：首先把1~10000素数表打出来，然后用bfs搜一遍，记录下步骤即可。

可能唯一有技巧性地就是从队列中取出一个数如何遍历改变，请看代码：

#include<iostream>#include<cstring>#include<queue>using namespace std;#define maxn 10000int s[10005],vis[10005],ans;struct node{    int x,step;};void Init(){    memset(s,0,sizeof(s));    int p=0;    for(int i=2;i*i<=maxn;i++)      if(s[i]==0)        for(int j=i*i;j<=maxn;j+=i)          s[j]=1;         }int bfs(int x){    if(x==ans)      return 0;    queue<node> q;    memset(vis,0,sizeof(vis));    q.push((node){x,0});        //将x对应的结构体压入队列     vis[x]=1;    while(!q.empty())    {        node a=q.front();        q.pop();        for(int i=1;i<=1000;i*=10)        {            for(int j=0;j<10;j++)            {                if(i==1000&&j==0)            //首位不能改成0                   continue;                int t=a.x%i+j*i+a.x/(i*10)*(i*10);        //改变把a.x上面第i位数改成j                 if(t==ans)                  return a.step+1;                if(s[t]==0)             //改动之后的t要为素数                 {                    if(vis[t])          //并且未被访问                       continue;                    vis[t]=1;                    q.push((node){t,a.step+1});      //压入队列                 }            }        }    }     return -1;}int main(){    Init();    int t,x;    cin>>t;    while(t--)    {        cin>>x>>ans;        cout<<bfs(x)<<endl;     }    return 0;}