看题目。。。

看解题报告。。。

敲。。。。

题意转自：優YoU  http://user.qzone.qq.com/289065406/blog/1303623014

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

Prime Path

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9564		Accepted: 5497

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.

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).

One line for each case, either with a number stating the minimal cost or containing the word Impossible.

3
1033 8179
1373 8017
1033 1033

6
7
0

我用的ＳＴＬ的ｑｕｅｕｅ和ＢＦＳ代码如下：

  1 #include<stdio.h>
  2 #include<string.h>
  3 #include<stdlib.h>
  4 #include<queue>
  5 using namespace std;
  6 queue<struct node>que;
  7
  8 struct node
  9 {
 10     int prime;
 11     int step;
 12 };
 13
 14 int n,m;
 15 int vis[15000];
 16 int   prim(int digit)
 17 {
 18     if(digit==2 || digit==3)
 19         return 1;
 20     else if(digit<=1 || digit%2==0)
 21         return 0;
 22     else if(digit>3)
 23     {
 24         for(int i=3; i*i<=digit; i+=2)
 25             if(digit%i==0)
 26                 return 0;
 27         return 1;
 28     }
 29 }
 30
 31 int  bfs()
 32 {
 33     int i;
 34     struct node k;
 35     while(!que.empty())
 36         que.pop();
 37     k.prime=n;
 38     k.step=0;
 39     que.push(k);
 40     vis[n]=1;
 41     while(!que.empty())
 42     {
 43         struct node s=que.front();
 44         que.pop();
 45         if(s.prime==m)
 46         {
 47             printf("%d\n",s.step);
 48             return 0;
 49         }
 50         int x=s.prime%10;
 51         int y=(s.prime/10)%10;
 52         for(i=1; i<=9; i+=2)
 53         {
 54             int a=(s.prime/10)*10+i;
 55             if(a!=s.prime&&!vis[a]&&prim(a))
 56             {
 57                 vis[a]=1;
 58                 k.prime=a;
 59                 k.step=s.step+1;
 60                 que.push(k);
 61             }
 62         }
 63         for(i=0; i<=9; i++)   //枚举十位
 64         {
 65             int a=(s.prime/100)*100+i*10+x;
 66             if(a!=s.prime && !vis[a] &&prim(a))
 67             {
 68                 vis[a]=1;
 69                 k.prime=a;
 70                 k.step=s.step+1;
 71                 que.push(k);
 72             }
 73         }
 74
 75         for(i=0; i<=9; i++) //枚举百位
 76         {
 77             int a=(s.prime/1000)*1000+i*100+y*10+x;
 78             if(a!=s.prime&&vis[a]==0&&prim(a))
 79             {
 80                 vis[a]=1;
 81                 k.prime=a;
 82                 k.step=s.step+1;
 83                 que.push(k);
 84             }
 85         }
 86         for(i=1; i<=9; i++)   //枚举的千位,保证四位数，千位最少为1
 87         {
 88             int a=s.prime%1000+i*1000;
 89             if(a!=s.prime && !vis[a] && prim(a))
 90             {
 91                 vis[a]=1;
 92                 k.prime=a;
 93                 k.step=s.step+1;
 94                 que.push(k);
 95             }
 96         }
 97     }
 98     printf("Impossible\n");
 99     return 0;
100 }
101 int  main()
102 {
103     int t;
104     scanf("%d",&t);
105     while(t--)
106     {
107         scanf("%d %d",&n,&m);
108         memset(vis,0,sizeof(vis));
109         bfs();
110     }
111     return 0;
112 }