Hotaru's problem

Posted 2020-08-26 神犇(shenben)

Hotaru\'s problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 3314    Accepted Submission(s): 1101

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

 

 

Author

UESTC

 

 

Source

2015 Multi-University Training Contest 7

 

//在manachar的基础上，枚举回文串的中心，再找第三部分。 
#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
int read(){
    register int x=0;bool f=1;
    register char ch=getchar();
    while(ch<\'0\'||ch>\'9\'){if(ch==\'-\')f=0;ch=getchar();}
    while(ch>=\'0\'&&ch<=\'9\'){x=x*10+ch-\'0\';ch=getchar();}
    return f?x:-x;
}
const int N=3e5+10;
int n,ans,cas,l,T,s[N],S[N],p[N];
void manacher(){
    int id=0,mx=-1;
    for(int i=1;i<l;i++){
        if(id+mx>i) p[i]=min(p[id*2-i],id+mx-i);
        while(i-p[i]>=0&&i+p[i]<=l&&S[i-p[i]]==S[i+p[i]]) p[i]++;
        if(id+mx<i+p[i]) id=i,mx=p[i];
    }
}
void init(){
    l=0;memset(p,0,sizeof p);
    for(int i=0;i<n;i++) S[++l]=-1,S[++l]=s[i];
    S[++l]=-1; 
}
int main(){
    for(T=read(),cas=1;ans=0,cas<=T;cas++){
        n=read();
        for(int i=0;i<n;i++) s[i]=read();
        init();manacher();
        for(int i=1;i<=n*2+1;i+=2){
            for(int j=i+p[i]-1;j-i>ans;j-=2){
                if(j-i+1<=p[j]){
                    ans=max(ans,j-i);
                    break;
                }
            }
        }
        printf("Case #%d: %d\\n",cas,ans/2*3);
    }
    return 0;
}