hdu5371 Hotaru's problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2189    Accepted Submission(s): 774

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

这题能够用Manacher算法做。由于题目要找的是三段（第一段和第二段对称，第二段和第三段对称）。事实上就是两个连在一起的回文串，我们能够先用Manacher算法初始化各个点的p[i]值（即能够向右延伸的最大距离。包含本身，这时已经增加了-1取代算法中的'#',-2取代算法中的'$'），然后对于每一个i。枚举j（j属于1~p[i]-1),假设i+j-p[i+j]+1<=i，那么说明i。j能够分别作为第一、二段的点和第二、三段的点）。
这里有个优化，由于枚举时满足条件的仅仅有'#'（即'-1’），所以我们能够使i，j每次变化2.

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
using namespace std;
#define maxn 100060
int a[maxn],b[2*maxn],p[2*maxn];
int main()
{
    int n,m,i,j,T,mx,idx,maxx,num1=0;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        for(i=0;i<n;i++){
            scanf("%d",&a[i]);
        }
        if(n<3){
            printf("0
");continue;
        }
        b[0]=-2;
        b[1]=-1;
        for(i=0;i<n;i++){
            b[i*2+2]=a[i];
            b[i*2+3]=-1;
        }
        n=2*n+2;mx=0;
        for(i=0;i<n;i++){
            if(i<mx){
                p[i]=min(p[idx*2-i],mx-i);
            }
            else p[i]=1;
            while(b[i-p[i]]==b[i+p[i]]){
                p[i]++;
            }
            if(mx<i+p[i]){
                mx=i+p[i];
                idx=i;
            }
        }
        maxx=0;
        for(i=3;i<n;i+=2){
            for(j=p[i]-1;j>=1;j-=2){
                if(j<maxx)break;
                if(i+j-p[i+j]+1<=i){
                    maxx=max(maxx,j);break;
                }
            }

        }
        num1++;
        printf("Case #%d: ",num1);
        printf("%d
",maxx/2*3);
    }
    return 0;
}