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".

真是被出题人打败了……给两个串,要判断用后缀树还是后缀自动机做……

其实就是模拟了

不想讲太多,直接看代码吧

#include<bits/stdc++.h>
#include<iostream>
using namespace std;
int d[10000],i,x,y;
char s[1000],c[1000];
int main(){
scanf("%s\n%s",s,c);
for(i=0;i<strlen(s);i++){
if(c[x]==s[i])x++;
if(x==strlen(c))break;
}
if(x==strlen(c)){
printf("automaton");
return 0;
}
for(i=0;i<strlen(s);i++)
d[s[i]]++;
for(i=0;i<strlen(c);i++){
if(d[c[i]]>0)d[c[i]]--;
else y=1;
}
if(y==0&&strlen(s)==strlen(c))printf("array");
else if(y==0)printf("both");
else printf("need tree");
return 0;
}

cf448B Suffix Structures的更多相关文章

  1. Codeforces Round #256 (Div. 2) B. Suffix Structures(模拟)

    题目链接:http://codeforces.com/contest/448/problem/B --------------------------------------------------- ...

  2. Codeforces Round #256 (Div. 2) B Suffix Structures

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

  3. Codeforces Round #256 (Div. 2) B (448B) Suffix Structures

    题意就是将第一个字符串转化为第二个字符串,支持两个操作.一个是删除,一个是更换字符位置. 简单的字符串操作!. AC代码例如以下: #include<iostream> #include& ...

  4. B. Suffix Structures 模拟吧,情况比較多要想周全

    这道题须要考虑的情况比較多,flag1表示情况是:b数组里有的字母而a里没有和b里面的同一个字母个数比a里面的多 flag2表示情况:b里面的左右字母能不能在a中同等顺序的存在 flag3表示情况:a ...

  5. codeforces 448B. Suffix Structures 解题报告

    题目链接:http://codeforces.com/problemset/problem/448/B 题目意思:给出两种操作automaton:可以删除字符串中任意一个字符: array:交换字符串 ...

  6. Codeforces Round #256 (Div. 2/B)/Codeforces448B_Suffix Structures(字符串处理)

    解题报告 四种情况相应以下四组数据. 给两字符串,推断第一个字符串是怎么变到第二个字符串. automaton 去掉随意字符后成功转换 array 改变随意两字符后成功转换 再者是两个都有和两个都没有 ...

  7. 【组队赛三】-C cf448B

    Suffix Structures Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u Submit S ...

  8. Codeforces Round #256 (Div. 2)

    A - Rewards 水题,把a累加,然后向上取整(double)a/5,把b累加,然后向上取整(double)b/10,然后判断a+b是不是大于n即可 #include <iostream& ...

  9. CodeForces 448

    A:Rewards: 题目链接:http://codeforces.com/problemset/problem/448/A 题意:Bizon有a1个一等奖奖杯,a2个二等奖奖杯,a3个三等奖奖杯,b ...

随机推荐

  1. Java:单例模式的七种写法[转]

    第一种(懒汉,线程不安全):  1 public class Singleton {   2     private static Singleton instance;   3     privat ...

  2. servlet下载,解决文件名中有中文下载路径出现乱码不能正常下载问题

    方法很多种,我只试用了两种. 主页面JSP中引入下载功能所需的js文件.引入的时候设置编码格式例如 <script type="text/javascript" charse ...

  3. JavaScript运算符有哪些

    JavaScript中的运算符有很多,主要分为算术运算符,等同全同运算符,比较运算符,字符串运算符,逻辑运算符,赋值运算符等.这些运算符都有一些属于自己的运算规则,下面就为大家介绍一下JavaScri ...

  4. Citrix 服务器虚拟化之十 Xenserver高可用性HA

    Citrix 服务器虚拟化之十 Xenserver高可用性HA HA是一套全自动功能设计,规划.它可以安全地恢复出现问题的XenServe 主机.例如物理破坏网络或主机的硬件故障,HA可确保无需任何人 ...

  5. oracle的启动和关闭

    一.sql*plus方式:      用sql*plus来连接到Oracle Sqlplus /nolog 是以不连接数据库的方式启动sql*plus Connect /as sysdba 是以DBA ...

  6. SELECT--UNION,UNION ALL,MINUS, INTERSECT,EXISTS

    SELECT--UNION,UNION ALL,MINUS, INTERSECT返回两个查询结果的集合操作,两个查询结果集必须字段相同.UNION和UNION ALL并集操作,UNION并集后去掉重复 ...

  7. C#基础:集合

        C#中的数组实现为 System.Array 类的实例,它们只是集合类(Collection Classes)中的一种类型.集合类一般用于处理对象列表,其功能比简单数组要多,功能大多是通过实现 ...

  8. 学习iOS必须知道的[转载]

    part1 : http://www.cocoachina.com/ios/20150608/12052.html part2 : http://www.cocoachina.com/ios/2015 ...

  9. React数据传递

    React基础概念 React是基于组件化的开发,通过组件的组合,让web应用能够实现桌面应用的效果. React更有利于单页应用的开发. 并非MVC框架,只能算是V 具有单项数据流的特点 优势:代码 ...

  10. 2.IKAnalyzer 中文分词器配置和使用

    一.配置 IKAnalyzer 中文分词器配置,简单,超简单. IKAnalyzer 中文分词器下载,注意版本问题,貌似出现向下不兼容的问题,solr的客户端界面Logging会提示错误. 给出我配置 ...