Prime Path（POJ 3126 BFS）

Prime Path
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 15325   Accepted: 8634

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
单纯的将所有情况都搜一遍
  1 #include <iostream>
  2 #include <cstring>
  3 #include <algorithm>
  4 #include <queue>
  5 using namespace std;
  6 bool prime[10000];
  7 bool vis[10000];
  8 int n[15];
  9 int a,b;
 10 struct node
 11 {
 12     int num,t;
 13 };
 14 void init()
 15 {
 16     int i,j;
 17     memset(prime,0,sizeof(prime));
 18     for(i=1000;i<9999;i++)
 19     {
 20         for(j=2;j<i;j++)
 21         {
 22             if(i%j==0)
 23                 break;
 24         }
 25         if(j==i)
 26         {
 27             prime[i]=1;    //1代表质数
 28             //cout<<i<<endl;
 29         }
 30     }
 31     for(i=0;i<10;i++)
 32         n[i]=i;
 33     n[10]=1,n[11]=3,n[12]=7,n[13]=9;
 34     return;
 35 }
 36 int bfs()
 37 {
 38     int i,j;
 39     queue<node> Q;
 40     node tem,u;
 41     int k;
 42     tem.num=a,tem.t=0;
 43     memset(vis,0,sizeof(vis));
 44     Q.push(tem);
 45     while(!Q.empty())
 46     {
 47         tem=Q.front();
 48         Q.pop();
 49         //cout<<tem.num<<endl;
 50         /*if(tem.num==3733)
 51             printf("adsfadsf\n");*/
 52         if(tem.num==b)
 53             return tem.t;
 54         for(i=10;i<=13;i++)
 55         {
 56             if(tem.num%10==n[i])
 57                 continue;
 58             u.num=tem.num/10;
 59             u.num=u.num*10+n[i];
 60             //cout<<u.num<<" "<<prime[u.num]<<" "<<vis[u.num]<<endl;
 61             if(vis[u.num]==0&&prime[u.num])
 62             {
 63                 u.t=tem.t+1;
 64                 vis[u.num]=1;
 65                 Q.push(u);
 66             }
 67         }
 68         for(i=0;i<=9;i++)
 69         {
 70             k=(tem.num/10)%10;
 71             if(k==n[i])
 72                 continue;
 73             k=tem.num%10;
 74             u.num=((tem.num/100)*10+n[i])*10+k;
 75             if(vis[u.num]==0&&prime[u.num])
 76             {
 77                 u.t=tem.t+1;
 78                 vis[u.num]=1;
 79                 Q.push(u);
 80             }
 81         }
 82         for(i=0;i<=9;i++)
 83         {
 84             k=(tem.num/100)%10;
 85             if(k==n[i])
 86                 continue;
 87             k=tem.num%100;
 88             u.num=((tem.num/1000)*10+n[i])*100+k;
 89             if(vis[u.num]==0&&prime[u.num])
 90             {
 91                 u.t=tem.t+1;
 92                 vis[u.num]=1;
 93                 Q.push(u);
 94             }
 95         }
 96         for(i=1;i<=9;i++)
 97         {
 98             k=tem.num/1000;
 99             if(k==n[i])
100                 continue;
101             //cout<<k<<" "<<n[i]<<endl;
102             k=tem.num%1000;
103             u.num=n[i]*1000+k;
104         //    cout<<u.num<<endl;
105             if(vis[u.num]==0&&prime[u.num])
106             {
107                 u.t=tem.t+1;
108                 vis[u.num]=1;
109                 //cout<<u.num<<endl;
110                 Q.push(u);
111             }
112         }
113         //break;
114     }
115     return -1;
116 }    
117 int main()
118 {
119     int T,ans;
120     init();
121     freopen("in.txt","r",stdin);
122     //freopen("ou.txt","w",stdin);
123     scanf("%d",&T);
124     while(T--)
125     {
126         scanf("%d%d",&a,&b);
127         ans=bfs();
128         if(ans==-1)    printf("Impossible\n");
129         else printf("%d\n",ans);
130     }
131 }

 

posted @ 2016-02-13 11:35  御心飞行  阅读(132)  评论(0)  编辑  收藏  举报