Prime Path（poj 3126)

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

题目大意：问输入的第一个数金过几次变换可以得到第二个数；
变换时，每次只能改变一个数字；
经过变换得到的数字必须是素数；
不能完成输出Impossible;

 1 #include<iostream>

 2 #include<cstring>

 3 #include<cstdio>

 4 #include<queue>

 5 using namespace std;

 6 bool pr[10003],vis[10003];

 7 int a,b,t,i,j;

 8 struct node

 9 {

10     int a,step;

11 }p,q;

12 void pri()

13 {

14     memset(pr,-1,sizeof(pr));

15     pr[0]=pr[1]=0;

16     for(i=2; i<10003; i++)

17     {

18         if(pr[i])

19             for(j=2*i; j<10003; j+=i)

20                 pr[j]=0;

21     }

22 }

23 int change(int x,int i,int j)

24 {//方便改变数字的每一位，x是原数字，i代表第几位i=1是个位，j是改编成几（0————9，千位不能为0）

25     if(i==1) return (x/10)*10+j;

26     else if(i==2) return (x/100)*100+x%10+j*10;

27     else if(i==3) return (x/1000)*1000+x%100+j*100;

28     else if(i==4) return (x%1000)+j*1000;

29 }

30 void bfs()//简单bfs

31 {

32     queue<node>que;

33     p.a=a;

34     p.step=0;

35     vis[a]=1;

36     que.push(p);

37     while(!que.empty())

38     {

39         p=que.front();

40         que.pop();

41         q.step=p.step+1;

42         for(i=1; i<5; i++)

43             for(j=0; j<10; j++)

44             {

45                 if(i==4&&j==0)

46                     continue;

47                 q.a=change(p.a,i,j);

48                 if(q.a==b)

49                 {

50                     printf("%d\n",q.step);

51                     return;

52                 }

53                 if(pr[q.a]&&!vis[q.a])

54                 {

55                     que.push(q);

56                     vis[q.a]=1;

57                 }

58             }

59     }

60     printf("Impossible\n");

61 }

62 int main()

63 {

64     pri();//素数筛初始化

65     scanf("%d",&t);

66     while(t--)

67     {

68         memset(vis,0,sizeof(vis));//初始化

69         scanf("%d %d",&a,&b);

70         if(a==b){printf("0\n");continue;}//a==b情况单独处理；

71         bfs();

72     }

73     return 0;

74 }

View Code