POJ3126 素数之路

题目内容 
Prime Path
Time Limit: 1000MS        Memory Limit: 65536K
Total Submissions: 6432        Accepted: 3673

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

Source
Northwestern Europe 2006

题目很长，大概意思是给你a、b两个四位素数，每一步可以将a的一个数字变成其他数字（0-9），但要求每次变换得到的结果都是素数且a始终为四位数。编程求解最少的变换次数。

依旧DFS，看准a始终为四位数就没什么难度。

我的基本方法：每次都把a拆成四个一位数，对这四个数逐个加\减1-9同时保证其<10或>=0。

View Code 
  1 #include<cstdio>
  2 #include<cstring>
  3 #include<queue>
  4 #define MAXN 10001
  5 using namespace std;
  6 bool prime[MAXN],used[MAXN];
  7 long step[MAXN];
  8 long N,beg,tot;
  9 void Primemaker()
 10 {
 11     long i,j;
 12     memset(prime,true,sizeof(prime));
 13     for (i=2;i<=100;i++)
 14     {
 15         if (prime[i])
 16         {
 17             for (j=i*i;j<MAXN;j+=i)
 18                 prime[j]=false;
 19         }
 20     }
 21     prime[0]=prime[1]=false;
 22 }
 23 void change(long n[],long x)
 24 {
 25     n[0]=x/1000;
 26     x%=1000;
 27     n[1]=x/100;
 28     x%=100;
 29     n[2]=x/10;
 30     x%=10;
 31     n[3]=x;
 32 }
 33 long back(long n[])
 34 {
 35     return n[0]*1000+n[1]*100+n[2]*10+n[3];
 36 }
 37 long bfs()
 38 {
 39     queue<long>q;
 40     if (beg==tot)
 41         return 0;
 42     step[beg]=0;
 43     long oper,t;
 44     q.push(beg);
 45     used[beg]=true;
 46     while(!q.empty())
 47     {
 48         oper=q.front();
 49         q.pop();
 50         if(oper==tot)
 51             return step[oper];
 52         long de[4];
 53         change(de,oper);    //分散存储oper
 54         for (long i=1;i!=10;i++)        //加
 55         {
 56                     for(long j=0;j!=4;j++)
 57                     {
 58                         if(de[j]+i<10)
 59                         {
 60                             de[j]+=i;
 61                             t=back(de);
 62                             if(!used[t]&&prime[t]&&t>1000)
 63                             {
 64                                 used[t]=true;
 65                                 step[t]=step[oper]+1;
 66                                 if(t==tot)
 67                                     return step[t];
 68                                 q.push(t);
 69                             }
 70                             de[j]-=i;
 71                         }
 72                     }
 73                 }
 74                 for(long i=-1;i!=-10;i--)        //减
 75                 {
 76                     for(long j=0;j!=4;j++)
 77                     {
 78                         if(de[j]+i>=0)
 79                         {
 80                             de[j]+=i;
 81                             t=back(de);
 82                             if(!used[t]&&prime[t]&&t>1000)
 83                             {
 84                                 used[t]=true;
 85                                 step[t]=step[oper]+1;
 86                                 if(t==tot)
 87                                     return step[t];
 88                                 q.push(t);
 89                             }
 90                             de[j]-=i;
 91                         }
 92                     }
 93                 }
 94     }
 95     return -1;
 96 }
 97 void init()
 98 {
 99     memset(used,false,sizeof(used));
100 }
101 int main(void)
102 {
103     Primemaker();
104     scanf("%ld",&N);
105     for(long i=0;i!=N;++i)
106     {
107         scanf("%ld%ld",&beg,&tot);
108         init();
109         long ans=bfs();
110         if (ans<0)
111             printf("Impossible\n");
112         else
113             printf("%ld\n",ans);
114     }
115     return 0;
116 }


应用了STL的queue，G++编译，240K 16MS

posted on 2011-10-30 18:50  SilVeRyELF  阅读(260)  评论(0)    收藏  举报

导航