双向广搜 POJ 3126 Prime Path

 

POJ 3126  Prime Path

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 16204		Accepted: 9153

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

大致题意：

给定两个四位素数a  b，要求把a变换到b

变换的过程要保证  每次变换出来的数都是一个 四位素数，而且当前这步的变换所得的素数  与  前一步得到的素数  只能有一个位不同，而且每步得到的素数都不能重复。

求从a到b最少需要的变换次数。无法变换则输出Impossible

注意：双向广搜是在一个队列中实现的，只不过是交替进行罢了！

 #include<iostream>
 using namespace std;
 #include<cstdio>
 #include<cstring>
 #include<cmath>
 #include<cstdlib>
 #include<queue>
 #define N 10000
 struct prime{
     int c[];
     int flag;
 };
 int dis[N];
 int visit[N];
 queue<prime>que;
 int js(int *m)
 {
     return (m[]*+m[]*+m[]*+m[]*);
 }
 bool is_prime(int l)
 {
     bool flag=true;
     for(int i=;i<=sqrt(l);++i)
     {
         if(l%i==)
         {
             flag=false;
             break;
         }
     }
     return flag;
 }
 int bfs()
 {
     while(!que.empty())
     {
         prime x=que.front();
         que.pop();
         int now=js(x.c);
         for(int i=;i<=;++i)
         {
                 prime nx=x;
                 nx.c[]=i;
                 int shu=js(nx.c);
                 if(!visit[shu]&&is_prime(shu))
                 {
                     visit[shu]=x.flag;
                     nx.flag=x.flag;
                     que.push(nx);
                     dis[shu]=dis[now]+;
                 }
                 else if(visit[shu]&&visit[shu]!=x.flag)
                 {
                     return dis[now]+dis[shu]+;
                 }
         }
         for(int j=;j<=;++j)
         {
             for(int i=;i<=;++i)
           {
                 prime nx=x;
                 nx.c[j]=i;
                 int shu=js(nx.c);
                 if(!visit[shu]&&is_prime(shu))
                 {
                     visit[shu]=x.flag;
                     nx.flag=x.flag;
                     que.push(nx);
                     dis[shu]=dis[now]+;
                 }
                 else if(visit[shu]&&visit[shu]!=x.flag)
                 {
                     return dis[now]+dis[shu]+;
                 }
             }

         }
     }
     return -;
 }
 int main()
 {
     int tex;
     scanf("%d",&tex);
     char a[];
     while(tex--)
     {
         while(!que.empty()) que.pop();
         memset(dis,,sizeof(dis));
         memset(visit,,sizeof(visit));
         scanf("%s",a);
         que.push(prime{a[]-'',a[]-'',a[]-'',a[]-'',});
         int p=(a[]-'')*+(a[]-'')*+(a[]-'')*+(a[]-'');
         visit[p]=;dis[p]=;
         int q=p;
         scanf("%s",a);
         que.push(prime{a[]-'',a[]-'',a[]-'',a[]-'',});
         p=(a[]-'')*+(a[]-'')*+(a[]-'')*+(a[]-'');
         visit[p]=;dis[p]=;
         if(q==p)
         {
             printf("0\n");continue;
         }
         int temp=bfs();
         if(temp==-) printf("Impossible\n");
         else printf("%d\n",temp);
      }
     return ;
 }