uva 129 Krypton Factor

 

Description

 

You have been employed by the organisers of a Super Krypton Factor Contest in which contestants have very high mental and physical abilities. In one section of the contest the contestants are tested on their ability to recall a sequence of characters which has been read to them by the Quiz Master. Many of the contestants are very good at recognising patterns. Therefore, in order to add some difficulty to this test, the organisers have decided that sequences containing certain types of repeated subsequences should not be used. However, they do not wish to remove all subsequences that are repeated, since in that case no single character could be repeated. This in itself would make the problem too easy for the contestants. Instead it is decided to eliminate all sequences containing an occurrence of two adjoining identical subsequences. Sequences containing such an occurrence will be called ``easy''. Other sequences will be called ``hard''.

For example, the sequence ABACBCBAD is easy, since it contains an adjoining repetition of the subsequence CB. Other examples of easy sequences are:

 

• BB
• ABCDACABCAB
• ABCDABCD

Some examples of hard sequences are:

 

• D
• DC
• ABDAB
• CBABCBA

Input and Output

In order to provide the Quiz Master with a potentially unlimited source of questions you are asked to write a program that will read input lines that contain integers n and L (in that order), where n > 0 and L is in the range  , and for each input line prints out the nth hard sequence (composed of letters drawn from the first L letters in the alphabet), in increasing alphabetical order (alphabetical ordering here corresponds to the normal ordering encountered in a dictionary), followed (on the next line) by the length of that sequence. The first sequence in this ordering is A. You may assume that for given n and L there do exist at least n hard sequences.

For example, with L = 3, the first 7 hard sequences are:

 

A 
AB 
ABA 
ABAC 
ABACA 
ABACAB 
ABACABA

As each sequence is potentially very long, split it into groups of four (4) characters separated by a space. If there are more than 16 such groups, please start a new line for the 17th group.

Therefore, if the integers 7 and 3 appear on an input line, the output lines produced should be

 

ABAC ABA
7

Input is terminated by a line containing two zeroes. Your program may assume a maximum sequence length of 80.

 

Sample Input

 

30 3
0 0

 

Sample Output

 

ABAC ABCA CBAB CABA CABC ACBA CABA
28

难点在于是否存在相邻重复子串的判断，类似于八皇后问题，我们默认已求得的串都是困难的串，即不存在相邻重复子串。
那么我们只需要考虑新加入的字母对于原串的影响，这样一层层地求，正好满足条件：已求得的串都是困难的串
对于dfs，由于相同的字母是允许存在的，因此将其改为二维数组，记录字符位与字符值。

#include <cstdio>
#include <iostream>
#include <cstring>
using namespace std;
int a[100];
int book[100][30];
int n,l;
bool cm(int b[],int cur,int i)
{
    for(int j=cur;j>=cur-i;j--)
    {
        if(b[j]!=b[j-i-1]) return false;
    }
    return true;
}
inline bool judge(int a[],int cur,int x)
{
    int b[100];memcpy(b,a,cur*sizeof(int));
    b[cur]=x;
    for(int i=0;cur-(i+1)*2>=0;i++)
    {
        if(cm(b,cur,i)) return false;
    }
    return true;
}
inline void Print(int len)
{
    int g=0;
    for(int i=1;i<len;i++)
    {
        printf("%c",a[i]-1+'A');
        if(i==64&&i!=len-1) puts("");
        else
        if(i%4==0&&i!=len-1)
        {
            printf(" ");
        }
    }
    printf("\n%d\n",len-1);
}
int count;bool flag;
void dfs(int cur,int pre)
{
    if(flag) return;
    if(pre!=-1) count++;
    if(count==n&&!flag)
    {
        Print(cur);
        flag=true;
        return;
    }
    for(int i=1;i<=l;i++)
    {
       if(book[cur][i]) continue;
       if(cur==1)
       {
           a[cur]=i;
           book[cur][i]=1;
           dfs(cur+1,i);
           book[cur][i]=0;
       }
       else
       {
           if(pre==i) continue;
           if(judge(a,cur,i))
           {
               a[cur]=i;
               book[cur][i]=1;
               dfs(cur+1,i);
               book[cur][i]=0;
           }
       }
     }
}
int main()
{
    while(~scanf("%d%d",&n,&l),n+l)
    {
        memset(book,0,sizeof(book));
        count=0;flag=false;
        dfs(1,-1);
    }
    return 0;
}