POJ 3126 Prime Path

Prime Path

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other)
Total Submission(s) : 7   Accepted Submission(s) : 3
Problem 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
31033 81791373 80171033 1033
 

Sample Output
670
 

Source
PKU
 
一道简单的BFS+素数读表：

  1 #include <iostream>
  2 #include <cstring>
  3 #include <cmath>
  4 #include <queue>
  5  
  6 using namespace std;
  7  
  8 int isprime[10005];
  9  
 10 struct nnd
 11 {
 12     int first;
 13     int second;
 14 };
 15  
 16 int a,e;
 17 int vis[10005];
 18  
 19 void dfs(queue<nnd> q)
 20 {
 21     while(!q.empty())
 22     {
 23         nnd t;
 24         nnd r;
 25         t=q.front();
 26         q.pop();
 27         //cout<<e<<endl;
 28         if(t.first==e)
 29         {
 30             //cout<<"hello";
 31             cout<<t.second<<endl;
 32             return ;
 33         }
 34  
 35         int k=t.first%1000;
 36         for(int i=1000;i<10000;i=i+1000)
 37         {
 38             if(isprime[k+i]&&!vis[k+i])
 39             {
 40                r.first=k+i;   r.second=t.second+1;
 41                q.push(r);
 42                vis[i+k]=1;
 43             }
 44         }
 45  
 46         int kk=t.first%100;
 47         int kkk=t.first/1000*1000;
 48         k=kk+kkk;
 49         for(int i=0;i<1000;i=i+100)
 50         {
 51             if(isprime[i+k]&&!vis[i+k])
 52             {
 53                 r.first=i+k;
 54                 r.second=t.second+1;
 55                 q.push(r);
 56                 vis[i+k]=1;
 57             }
 58         }
 59  
 60         kk=t.first%10;
 61         kkk=t.first/100*100;
 62         k=kk+kkk;
 63         for(int i=0;i<100;i=i+10)
 64         {
 65             if(isprime[i+k]&&!vis[i+k])
 66             {
 67                 r.first=i+k;
 68                 r.second=t.second+1;
 69                 q.push(r);
 70                 vis[i+k]=1;
 71             }
 72         }
 73  
 74         k=t.first/10*10;
 75         for(int i=1;i<10;i=i+2)
 76         {
 77             if(isprime[i+k]&&!vis[i+k])
 78             {
 79                 r.first=i+k;
 80                 r.second=t.second+1;
 81                 q.push(r);
 82                 vis[i+k]=1;
 83             }
 84         }
 85     }
 86  
 87     cout<<"Impossible"<<endl;
 88     return;
 89 }
 90  
 91  
 92 int main()
 93 {
 94     int n;
 95     cin>>n;
 96  
 97     for(int i=0;i<10005;i++)
 98     {
 99         isprime[i]=i;
100     }
101     for(int i=2;i*i<10010;i++)
102     {
103         if(isprime[i]!=0)
104         for(int j=2;j*i<10010;j++)
105         {
106             isprime[i*j]=0;
107         }
108     }
109  
110     isprime[1]=0;
111  
112  
113     for(int o=0;o<n;o++)
114     {
115         queue<nnd> q;
116         nnd a;
117         memset(vis,0,sizeof(vis));
118         cin>>a.first>>e;
119         a.second=0;
120         q.push(a);
121         vis[a.first]=1;
122  
123         dfs(q);
124     }
125     return 0;
126 }

 


posted @ 2013-04-10 23:38  码代码的猿猿  阅读(207)  评论(0)  编辑  收藏  举报