hdu5371 最长回文子串变形（Manacher算法）

http://acm.hdu.edu.cn/showproblem.php?pid=5371

Problem Description

Hotaru Ichijou recently is addicated to math problems. Now she is playing with N-sequence.
Let's define N-sequence, which is composed with three parts and satisfied with the following condition:
1. the first part is the same as the thrid part,
2. the first part and the second part are symmetrical.
for example, the sequence 2,3,4,4,3,2,2,3,4 is a N-sequence, which the first part 2,3,4 is the same as the thrid part 2,3,4, the first part 2,3,4 and the second part 4,3,2 are symmetrical.

Give you n positive intergers, your task is to find the largest continuous sub-sequence, which is N-sequence.

 

Input

There are multiple test cases. The first line of input contains an integer T（T<=20）, indicating the number of test cases. 

For each test case:

the first line of input contains a positive integer N（1<=N<=100000）, the length of a given sequence

the second line includes N non-negative integers ,each interger is no larger than  109  , descripting a sequence.

 

Output

Each case contains only one line. Each line should start with “Case #i: ”,with i implying the case number, followed by a integer, the largest length of N-sequence.

We guarantee that the sum of all answers is less than 800000.

 

Sample Input

   
   
   
   

    
    
    
    1
10
2 3 4 4 3 2 2 3 4 4
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    Case #1: 9
   
   
   
   
   
   
   
   


   
   
   
   

/**
hdu5371 最长回文子串变形（Manacher算法）
题目大意：找出一个字符串可以均分为三段，第一段和第三段相同，和第二段互为回文串
解题思路：利用Manacher算出每个位置的最长回文子串的长度，然后枚举即可
*/
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <iostream>
using namespace std;
const int maxn=210001;
int s[maxn],a[maxn];
int r[maxn],len;

void Manancher()
{
    int l=0;
    a[l++]=-2;
    a[l++]=-1;
    for(int i=0;i<len;i++)
    {
        a[l++]=s[i];
        a[l++]=-1;
    }
    a[l]=-3;
    int mx=0,id=0;
    for(int i=0;i<l;i++)
    {
        r[i]=mx>i?min(r[2*id-i],mx-i):1;
        while(a[i+r[i]]==a[i-r[i]])r[i]++;
        if(i+r[i]>mx)
        {
            mx=i+r[i];
            id=i;
        }
        //printf("%d ",r[i]);
    }
   // printf("\n");
}

int main()
{
    int T,tt=0;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&len);
        for(int i=0;i<len;i++)
        {
            scanf("%d",&s[i]);
        }
        Manancher();
        int ans=0;
        for(int i=1;i<=len*2+1;i+=2)
        {
            for(int j=i+r[i]-1;j-i>ans;j-=2)
            {
                if(j-i+1<=r[j])
                {
                    ans=max(ans,j-i);
                    break;
                }
            }
        }
        printf("Case #%d: %d\n",++tt,ans/2*3);
    }
    return 0;
}