559B - Equivalent Strings & 560D - Equivalent Strings

D. Equivalent Strings

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

Today on a lecture about strings Gerald learned a new definition of string equivalency. Two strings
a and b of equal length are called
equivalent in one of the two cases:

They are equal.
If we split string a into two halves of the same size
a1 and
a2, and string
b into two halves of the same size b1 and
b2, then one of the following is correct:

a1 is equivalent to
b1, and
a2 is equivalent to
b2
a1 is equivalent to
b2, and
a2 is equivalent to
b1

As a home task, the teacher gave two strings to his students and asked to determine if they are equivalent.

Gerald has already completed this home task. Now it's your turn!

Input
The first two lines of the input contain two strings given by the teacher. Each of them has the length from
1 to 200 000 and consists of lowercase English letters. The strings have the same length.

Output
Print "YES" (without the quotes), if these two strings are equivalent, and "NO" (without the quotes) otherwise.

Sample test(s)

Input

aaba
abaa

Output

YES

Input

aabb
abab

Output

NO

Note
In the first sample you should split the first string into strings "aa" and "ba", the second one — into strings "ab" and "aa".
"aa" is equivalent to "aa"; "ab" is equivalent to "ba" as "ab"
= "a" + "b", "ba" = "b" + "a".

In the second sample the first string can be splitted into strings "aa" and "bb", that are equivalent only to themselves. That's why string "aabb"
is equivalent only to itself and to string "bbaa".

分析：刚开始用的是string的substr函数，总是超时，我想可能是string有点慢吧，如果用普通字符串，可能就OK了

//卡在第89个样例超时
#include<bits/stdc++.h>
using namespace std;
bool solve(string s1,string s2 )
{
    if(s1==s2)return 1;
    if(s1.size()%2==1)return 0;
    int n=s1.size()>>1;
    string ss1=s1.substr(0,n);
    string ss11=s1.substr(n,n);
    string ss2=s2.substr(0,n);
    string ss22=s2.substr(n,n);
    if((solve(ss1,ss2)&&solve(ss11,ss22))||(solve(ss1,ss22)&&solve(ss11,ss2)))return 1;
    return 0;

}
int main()
{
    string s1,s2;
    cin>>s1>>s2;
        int flag=0;
        if(s1==s2){
            cout<<"YES"<<endl;
            return 0;
        }
        if(s1.size()!=s2.size()){
            cout<<"NO"<<endl;
            return 0;
        }
        if(s1.size()%2==1){
            cout<<"NO"<<endl;
            return 0;
        }
       if(solve(s1,s2))cout<<"YES"<<endl;
       else cout<<"NO"<<endl;
    return 0;
}

然后实在没有耐心改了，看了一下题解，恍然大悟啊。每次字符串一分为二那么肯定是字典序小的和小的比较，大的和大的比较，那么我们就改变一下字符串顺序在比较就好了。

题解：

Let us note that "equivalence" described in the statements is actually
equivalence relation, it is reflexively, simmetrically and transitive. It is meant that set of all string is splits to equivalence classes. Let's find lexicographic minimal strings what is equivalent to first and to second given string. And then check if
its are equals.

It is remain find the lexicographic minimal strings what is equivalent to given. For instance we can do it such a way:

[code]String smallest(String s) {
    if (s.length() % 2 == 1) return s;
    String s1 = smallest(s.substring(0, s.length()/2));
    String s2 = smallest(s.substring(s.length()/2), s.length());
    if (s1 < s2) return s1 + s2;
    else return s2 + s1;
}

Every recursive call time works is O(n) (where
n is length of strings) and string splitten by two twice smaller strings. Therefore time of work this function is


, where
n is length of strings.

//234ms
#include<bits/stdc++.h>
using namespace std;
string smallest(string s)
{
    if(s.size()%2==1)return s;
    int n=s.size()>>1;
    string s1=smallest(s.substr(0,n));
    string s2=smallest(s.substr(n,n));
    if(s1<s2)return s1+s2;
    else return s2+s1;
}
int main()
{
    string a,b;
    while(cin>>a>>b){
        if(smallest(a)==smallest(b))printf("YES\n");
        else printf("NO\n");
    }
    return 0;
}