poj Prime Path BFS

Prime Path

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7550		Accepted: 4281

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

---

简单的BFS题，分别改变一个四位数的个、十、百、千位即可：
代码：

View Code

  1 #include <stdio.h>

  2 #include <iostream>

  3 #include <string.h>

  4 using namespace std;

  5 

  6 

  7 struct M{

  8     int num;

  9     int step;

 10 };

 11 

 12 bool prime(int x)

 13 {

 14     int i,j;

 15     for(i=2;i*i<=x;i++)

 16     {

 17         if(x%i==0) 

 18             return false;

 19     }

 20     return true;

 21 }

 22 

 23 int BFS(int a, int b)

 24 {

 25     M queue[100000];

 26     bool vis[100000]={false};

 27     int i,j;

 28     int head,tail;

 29     head=tail=0;

 30     queue[0].num=a;

 31     queue[tail++].step=0;

 32     while(head<tail)

 33     {

 34         M x=queue[head++];

 35         if(x.num==b)

 36         {

 37             return x.step;

 38         }

 39         int aft;

 40         for(i=1;i<=9;i++)        //改变个位

 41         {

 42             if(!vis[x.num/10*10+i] && prime(x.num/10*10+i)==true)

 43             {

 44                 vis[x.num/10*10+i]=true;

 45                 queue[tail].num=x.num/10*10+i;

 46                 queue[tail++].step=x.step+1;

 47                 //cout<<x.num/10*10+i<<"....."<<endl;

 48             }

 49         }

 50         for(i=0;i<=9;i++)        //改变十位    

 51         {

 52             aft=x.num%10+x.num/100*100+i*10;

 53             if(!vis[aft] && prime(aft))

 54             {

 55                 vis[aft]=true;

 56                 queue[tail].num=aft;

 57                 queue[tail++].step=x.step+1;

 58                 //cout<<aft<<"......"<<endl;

 59             }

 60         }

 61         for(i=0;i<=9;i++)        //改变百位

 62         {

 63             aft=x.num%100+x.num/1000*1000+i*100;

 64             if(!vis[aft] && prime(aft))

 65             {

 66                 vis[aft]=true;

 67                 queue[tail].num=aft;

 68                 queue[tail++].step=x.step+1;

 69                 //cout<<aft<<"....."<<endl;

 70             }

 71         }

 72         for(i=1;i<=9;i++)      //改变千位

 73         {

 74             aft=x.num%1000+i*1000;

 75             if(!vis[aft] && prime(aft))

 76             {

 77                 vis[aft]=true;

 78                 queue[tail].num=aft;

 79                 queue[tail++].step=x.step+1;

 80                 //cout<<aft<<"......"<<endl;

 81             }

 82         }

 83     }

 84     return 0;

 85 }

 86 

 87 int main()

 88 {

 89     int t;

 90     int i,j,k;

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

 92     while(t--)

 93     {

 94         int n;

 95         int m;

 96         scanf("%d%d",&n,&m);

 97         printf("%d\n",BFS(n,m));

 98     }

 99     return 0;

100 }