Magic Necklace

Time limit: - MS Space limitations: - KB

Benchmark description: 1s, 64m

Problem Description

Mother's Day is coming, little H ready to give her a special necklace.
This string can be seen as a necklace with lowercase letters, lowercase letters each represent a color.

To make this necklace, small H purchased two machines.
The first machine may generate all forms of palindromic sequence, the second machine can be connected to two palindromic sequence, and the second machine there is a special property:
If a suffix string and a string of prefix is exactly the same, then this can be repeated partially overlap. For example: aba and aca connected, the string may be generated abaaca or abaca.

Now given target style necklace, you need to ask how many times a second machine in order to generate this special necklace.

Input Format

How the input data lines, each a string representing the target necklace style.

Output Format

Multiple lines, each line represents the minimum number of answers to a second machine is required.

Sample input 1

abcdcba
abacada
ABCDEF

Sample output 1

0
2
5

Sample input 2

xuqeytcixfzpzvcacymqncdohedfyowmipplplkyrsaspjliczflordhlbckyiuqxkslntofajs
amjmaekzbnbwagotspirvjksendltyeeuswefpdcdmmhzomlvkrhtwidlybkvvvebqkmvednaxddeygghrvqfaxwjssvcphcrzeauwlowwdmhacpzbnihgmbypfsblvsyaugkcg

Sample output 2

65
118

prompt

Each test data input less than 5 lines

String length of each row is less 50000

1. Identify all drawn cart through palindromic substrings, each of the palindromic regarded as a sub-string segment
2. The question becomes minimal is selected from the range covering the entire line segment, i.e., the minimum interval coverage problems. Greed can be.

#include <stdio.h> 
#include <bits / STDC ++ H.>
 the using  namespace STD;
 char S [ 100010 ];
 int len [ 100010 ]; // palindromic sequence length 
int n-;
 struct the Node 
{ 
    int L, R & lt; 
} A [ 100010 ];
 BOOL CMP (the Node A, the Node B) // the left end greedy ordering 
{
     return Al < BL; 
} 
void the Init () 
{ 
    for ( int I = n-; I> = . 1 ; i-- )  
    {
        S [i <<1]=s[i];
        s[i<<1|1]='&';
    }
    n=n<<1|1;
    s[0]='%';
    s[1]='&';
    s[n+1]='^';
}
void manacher()
{
    int maxright=0,num=0;
    for(int i=1; i<=n; i++)
    {
        if(i<maxright)
            len[i]=min(len[num*2-i],maxright-i);
        else
            len[i]=1;
        while(s[i+len[i]]==s[i-len[i]])
        {
            len[i]++;
        }
        if(maxright<i+len[i])
        {
            maxright=i+len[i];
            num=i;
        }
    }
}
intmain () 
{ 
    the while (Scanf ( " % S " , S + . 1 ) = the EOF!) // starts within the operator 
    { 
        n- = strlen (S + . 1 ); 
        the Init (); 
        manacher (); 
        for ( int I = . 1 ; i <= n-; i ++ ) 
        { 
            a [i] .L = i-len [i] + . 1 ; // in length i is the center of the left 
            a [i] = i + len .r [i] - . 1 ; // at the center of the right length i is 
        } 
        Sort (a + . 1 , a + . 1 + n-, CMP);
        int r=0,ans=0,i=1;
        while(i<=n)
        {
            int sum=0;
            while(a[i].l-1<=r&&i<=n)
            {
                sum=max(sum,a[i].r);
                i++;
            }
            ans++;//现长
            r=sum;//覆盖完全
            if(r==n)
            {
                break;
            }
            //++ years; 
        } 
        Cout << years- 1 << endl; 
    } 
}

manacher --- magical necklace