武大OJ 622. Symmetrical

Description

         Cyy likes something symmetrical, and Han Move likes something circular. Han Move gave you a circular string whose length is N, i.e. the i-th character of the string is the neighbor of the ( (i-1) mod N )-th character and ( (i+1) mod N )-th character.

 

         As Cyy likes something symmetrical, you want to find out whether this string can be symmetrical. There’s an operation for you to make the string symmetrical. This operation is called “disconnect”. You can use “disconnect” only once to break the link between any two neighbor characters, making the circular string into a linear string. If you can use “disconnect” to make this string into a palindrome, this string can be considered to be symmetrical.

 

         Please tell Cyy whether this circular string is symmetrical. 

 

Input

The input file consists of multiple test cases.
For each test case, there’s only a string in a line to represent the circular string. The string only consists of lowercase characters. ( 1 <= N <= 65536 )
It’s guaranteed that the sum of N is not larger than 1000000.

Output

For each test case, output “Yes” in a line if the circular string is symmetrical, output “No” otherwise.

Sample Input

aaaaaa
abcde
ababab
aaaaba
aabbaa

Sample Output

Yes
No
No
No
Yes

一句话题意：循环字符串能否断开形成回文串
循环的东西，很明显要直接复制一遍
回文串，很明显manacher

如果原串长度len为奇数，则找到的回文数长度m必须>=len且m为奇数

如果原串长度len为偶数，则找到的回文数长度m必须>=len且m为偶数

 一开始总是超时，一直以为是判断输入结束不对，后来才发现是用了string，每次都用string生成！#a#b..,很耗时！

#include<iostream>
#include<stdio.h>
#include<math.h>
#include<stdlib.h>
#include<string.h> 
#include<algorithm>
using namespace std;

int p[300000];
bool manacher(char* s,int len){
    int length=0;
    int maxRight=1;
    int mid=1;
    int l=strlen(s);
    for(int i=1;i<l;++i)
    {
        if(i<maxRight)
        p[i]=min(p[2*mid-i],maxRight-i);
        else p[i]=1;
        
        while(s[i-p[i]]==s[i+p[i]]) p[i]++;
        
        if(p[i]+i>maxRight) 
        {
            maxRight=p[i]+i;
            mid=i;
        }
        length=p[i]-1;    
        if(len%2==0)
        {
            if(!(length<len||length%2==1))
            return true;
        }
        if(len%2==1)
        {
            if(!(length<len||length%2==0))
            return true;
        }
    }
    return false;

}

int main()
{
    int len;
    char s[300000];
    
    while(~scanf("%s",&s))
    {
        len=strlen(s);
        for(int i=len;i>=0;--i)
        s[i+len]=s[i];
    //    cout<<s<<endl;
        for(int i=2*len;i>=0;--i)
        {
            s[2*i+2]=s[i];
            s[2*i+1]='#';
        }
        
        s[0]='!';
    //    cout<<s<<endl;

        if(manacher(s,len)==true)
        cout<<"Yes"<<endl;
        else cout<<"No"<<endl;    
    }

}