pid=5371">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.
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.
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.
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：找出一个字符串能够均分为三段，第一段和第三段同样。和第二段互为回文串
解题思路：利用Manacher算出每一个位置的最长回文子串的长度，然后枚举就可以
*/
#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;
}