Suffix Structures - #256 (Div. 2) B (448B) 陷阱题

Bizon the Champion面对一个关于字符串转换的问题,需要使用Suffix数据结构(如Suffix Automaton或Suffix Array)来判断是否能将一个字符串转换成另一个不同的字符串。文章详细介绍了如何通过移除字符或交换字符顺序来实现转换,并提供了三种可能的解决方案:单独使用Suffix Automaton、单独使用Suffix Array或同时使用两者。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

B. Suffix Structures
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

Bizon the Champion isn't just a bison. He also is a favorite of the "Bizons" team.

At a competition the "Bizons" got the following problem: "You are given two distinct words (strings of English letters), s and t. You need to transform word s into word t". The task looked simple to the guys because they know the suffix data structures well. Bizon Senior loves suffix automaton. By applying it once to a string, he can remove from this string any single character. Bizon Middle knows suffix array well. By applying it once to a string, he can swap any two characters of this string. The guys do not know anything about the suffix tree, but it can help them do much more.

Bizon the Champion wonders whether the "Bizons" can solve the problem. Perhaps, the solution do not require both data structures. Find out whether the guys can solve the problem and if they can, how do they do it? Can they solve it either only with use of suffix automaton or only with use of suffix array or they need both structures? Note that any structure may be used an unlimited number of times, the structures may be used in any order.

Input

The first line contains a non-empty word s. The second line contains a non-empty word t. Words s and t are different. Each word consists only of lowercase English letters. Each word contains at most 100 letters.

Output

In the single line print the answer to the problem. Print "need tree" (without the quotes) if word s cannot be transformed into word teven with use of both suffix array and suffix automaton. Print "automaton" (without the quotes) if you need only the suffix automaton to solve the problem. Print "array" (without the quotes) if you need only the suffix array to solve the problem. Print "both" (without the quotes), if you need both data structures to solve the problem.

It's guaranteed that if you can solve the problem only with use of suffix array, then it is impossible to solve it only with use of suffix automaton. This is also true for suffix automaton.

Sample test(s)
input
automaton
tomat
output
automaton
input
array
arary
output
array
input
both
hot
output
both
input
need
tree
output
need tree
Note

In the third sample you can act like that: first transform "both" into "oth" by removing the first character using the suffix automaton and then make two swaps of the string using the suffix array and get "hot".


题意:automaton是去掉任意字符,array是改变字符顺序,问第一个字符串能否通过这两种方法变成第二个字符串,不能的话输出need tree,可以的话,如果只用到一个,输出那一个,两个都用到,输出both。

思路:陷阱就在比如aabaa,可以通过automaton变成aaaa。

AC代码如下:

#include<cstdio>
#include<cstring>
using namespace std;
int num1[30],num2[30],len1,len2;
char s1[1010],s2[1010];
bool solve2()
{ int i,k=0;
  for(i=0;i<len1;i++)
  { if(k>=len2)
     break;
    if(s1[i]==s2[k])
     k++;
  }
  if(k==len2)
   return true;
  else
   return false;
}
void solve()
{ int n,i,j,k;
  bool flag=true;
  scanf("%s",s1);
  len1=strlen(s1);
  for(i=0;i<len1;i++)
   num1[s1[i]-'a']++;
  scanf("%s",s2);
  len2=strlen(s2);
  for(i=0;i<len2;i++)
   num2[s2[i]-'a']++;
  for(i=0;i<=26;i++)
   if(num2[i]>num1[i])
    flag=false;
  if(solve2())
  { printf("automaton\n");
    return;
  }
  if(!flag)
  { printf("need tree\n");
    return;
  }
  if(len1==len2)
   printf("array\n");
  else
   printf("both\n");
}
int main()
{ solve();
}


评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值