B. Suffix Structures

life4711發表於2014-07-19
Struct

http://codeforces.com/contest/448/problem/B

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".

題目大意：

               給定兩個字串s和t，將s經過刪除字元或者交換字串中字元的順序使其變成t，如果只需要交換輸出arry如果只需要刪除輸出automation，如果需要兩種操作同時進行就輸出both，如果不能實現就輸出need tree。

解題思路：

               先統計兩個字串中各種字元的個數，如果對應t的字元有比s對應的少那麼need tree，匹配t如果s可以按順序匹配出來t那麼就只用刪除就可以了，如果s和t中對應字元都相等那麼就arry，否則both

#include <string.h>
#include <stdio.h>
#include <iostream>
#include <math.h>
using namespace std;
char s[105],t[105];
int a[27],b[27];
int main()
{
    memset(s,0,sizeof(s));
    memset(t,0,sizeof(t));
    memset(a,0,sizeof(a));
    memset(b,0,sizeof(b));
    while(~scanf("%s%s",s,t))
    {
        int n=strlen(s);
        int m=strlen(t);
        for(int i=0; i<m; i++)
            a[t[i]-'a']++;
        for(int i=0; i<n; i++)
            b[s[i]-'a']++;
        int flag=1;
        int flag1=1;
        for(int i=0; i<26; i++)
        {
            if(a[i]>b[i])
                flag1=0;
            if(a[i]!=b[i])
                flag=0;
        }
        if(m==n&&flag==1)
        {
            printf("array\n");
        }
        else if(flag1==0||n<m)
        {
            printf("need tree\n");
        }
        else
        {
            int j=0;
            for(int i=0; i<n; i++)
                if(s[i]==t[j])j++;
            if(j==m)
            {
                printf("automaton\n");
            }
            else
                printf("both\n");
        }
        memset(s,0,sizeof(s));
        memset(t,0,sizeof(t));
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
    }
    return 0;
}



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