Prime Path(素数筛选+bfs)

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9519		Accepted: 5458

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

Sample Output

6
7
0
题意：给定两个四位数，求从前一个数变到后一个数最少需要几步，改变的原则是每次只能改变某一位上的一个数，而且每次改变得到的必须是一个素数；

思路：将四位数以内的素数筛选出来，bfs时，枚举四位数的每一位上的每一个数；

  1 #include<stdio.h>
  2 #include<string.h>
  3 #include<queue>
  4 using namespace std;
  5 
  6 struct node
  7 {
  8     int num;
  9     int step;
 10 };
 11 queue<struct node>que;
 12 int n,m,flag;
 13 bool p[10010],vis[10010];
 14 
 15 //素数筛；
 16 void prime_search()
 17 {
 18     memset(p,1,sizeof(p));
 19     for(int i = 4; i <= 10000; i+=2)
 20         p[i] = 0;
 21     for(int i = 3; i <= 100; i++)
 22     {
 23         if(p[i])
 24         {
 25             for(int j = i+i; j <= 10000; j += i)
 26                 p[j] = 0;
 27         }
 28     }
 29 }
 30 
 31 int bfs()
 32 {
 33     while(!que.empty())
 34         que.pop();
 35     que.push((struct node){n,0});
 36     vis[n] = 1;
 37     int res,tmp,r,t,s;
 38     while(!que.empty())
 39     {
 40         struct node u = que.front();
 41         que.pop();
 42         if(u.num == m)
 43             return u.step;
 44             
 45         //枚举个位数
 46         r = u.num%10;
 47         for(int k = -9; k <= 9; k++)
 48         {
 49             t = r+k;
 50             if(t >= 0 && t <= 9)
 51             {
 52                 res = (u.num/10)*10+t;
 53                 if(p[res] && !vis[res])
 54                 {
 55                     que.push((struct node){res,u.step+1});
 56                     vis[res] = 1;
 57                 }
 58             }
 59         }
 60         
 61         //枚举十位数
 62         tmp = u.num/10;
 63         r = tmp%10;
 64         s = tmp/10;
 65         for(int k = -9; k <= 9; k++)
 66         {
 67             t = r+k;
 68             if(t >= 0 && t <= 9)
 69             {
 70                 res = (s*10+t)*10+u.num%10;
 71                 if(p[res] && !vis[res])
 72                 {
 73                     que.push((struct node){res,u.step+1});
 74                     vis[res] = 1;
 75                 }
 76             }
 77         }
 78         
 79         //枚举百位数
 80         int tmp = u.num/100;
 81         r = tmp%10;
 82         s = tmp/10;
 83         for(int k = -9; k <= 9; k++)
 84         {
 85             t = r+k;
 86             if(t >= 0 && t <= 9)
 87             {
 88                 res = (s*10+t)*100+u.num%100;
 89                 if(p[res] && !vis[res])
 90                 {
 91                     que.push((struct node){res,u.step+1});
 92                     vis[res] = 1;
 93                 }
 94             }
 95         }
 96 
 97         //枚举千位数
 98         r = u.num/1000;
 99         for(int k = -9; k <= 9; k++)
100         {
101             t = r+k;
102             if(t >0 && t <= 9)
103             {
104                 res = t*1000+u.num%1000;
105                 if(p[res] && !vis[res])
106                 {
107                     que.push((struct node){res,u.step+1});
108                     vis[res] = 1;
109                 }
110             }
111         }
112     }
113 }
114 int main()
115 {
116     int test;
117     prime_search();
118     scanf("%d",&test);
119     while(test--)
120     {
121         memset(vis,0,sizeof(vis));
122         scanf("%d %d",&n,&m);
123         int ans = bfs();
124         printf("%d\n",ans);
125     }
126     return 0;
127 }

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posted on 2013-08-20 10:29 straw_berry 阅读(257) 评论(0) 编辑收藏举报