POJ 3126 Prime Path （bfs+欧拉线性素数筛）

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
 
题意给你两个4位素数a b，让你将a每次改变一位数字，改变后的4位数还必须是素数，最少几步能变到b，输出步数，不能变到输出Impossible。
这题第一发T了...一看就是果然T，忘了加vis2这个数组标记那些数组被访问过了...不加的话因为队列及时pop了，会出现在两个素数之间来回跳的情况！
小插曲就是欧拉线性素数筛，自己还没背会2333333。
代码如下：

 #include <iostream>
 #include <cstdio>
 #include <queue>
 #include <algorithm>
 #include <cstring>
 using namespace std;
 #define inf 0x3f3f3f3f
 int prime[],a,b,ans;
 bool num[],vis[],vis2[];
 bool flag;
 int dig[];
 struct node
 {
     int x,stp;
 };
 void split (int x)//将x分成各个位
 {
     for (int i=;i<;++i)
     {
         dig[i]=x%;
         x/=;
     }
 }
 int getnum (int a,int b,int c,int d)//将4个数字组成一个四位数
 {
     return a+b*+c*+d*;
 }
 void getprime()//欧拉线性素数筛
 {
     memset(num,false,sizeof num);
     memset(vis,false,sizeof vis);
     memset(prime,,sizeof prime);
     int cnt=;
     for (int i=;i<;++i)
     {
         if (!vis[i])
         prime[cnt++]=i,num[i]=;
         for (int j=;j<cnt&&i*prime[j]<;++j)
         {
             vis[i*prime[j]]=;
             if (i%prime[j]==)
             break;
         }
     }
 }
 void bfs(node now)
 {
     queue<node>q;
     q.push(now);
     vis2[now.x]=true;
     if (now.x==b)
     {
         flag=true;
         ans=now.stp;
         return ;
     }
     while (!q.empty())
     {
         node frt=q.front();
         q.pop();
         if (frt.x==b)
         {
             flag=true;
             ans=frt.stp;
             return ;
         }
         split(frt.x);
         for (int i=;i<=;++i)//改个位
         {
             int temp=getnum(i,dig[],dig[],dig[]);
             if (temp==frt.x)
             continue;
             if (num[temp]&&!vis2[temp])
             {
                 node tp;
                 tp.x=temp;
                 tp.stp=frt.stp+;
                 vis2[temp]=true;
                 q.push(tp);
             }
         }
         for (int i=;i<=;++i)//改十位
         {
             int temp=getnum(dig[],i,dig[],dig[]);
             if (temp==frt.x)
             continue;
             if (num[temp]&&!vis2[temp])
             {
                 node tp;
                 tp.x=temp;
                 tp.stp=frt.stp+;
                 vis2[temp]=true;
                 q.push(tp);
             }
         }
         for (int i=;i<=;++i)//改百位
         {
             int temp=getnum(dig[],dig[],i,dig[]);
             if (temp==frt.x)
             continue;
             if (num[temp]&&!vis2[temp])
             {
                 node tp;
                 tp.x=temp;
                 tp.stp=frt.stp+;
                 vis2[temp]=true;
                 q.push(tp);
             }
         }
         for (int i=;i<=;++i)//改千位，注意是4位数，所以千位不为0，从一开始
         {
             int temp=getnum(dig[],dig[],dig[],i);
             if (temp==frt.x)
             continue;
             if (num[temp]&&!vis2[temp])
             {
                 node tp;
                 tp.x=temp;
                 tp.stp=frt.stp+;
                 vis2[temp]=true;
                 q.push(tp);
             }
         }
     }
     return ;
 }
 int main()
 {
     getprime();
     //freopen("de.txt","r",stdin);
     int t;
     scanf("%d",&t);
     while (t--)
     {
         scanf("%d%d",&a,&b);
         memset(vis2,false,sizeof vis2);
         ans=inf;
         node now;
         now.x=a,now.stp=;
         bfs(now);
         if (flag)
         printf("%d\n",ans);
         else
         printf("Impossible\n");
     }
     return ;
 }