HDU 5371 Hotaru's problem（manacher算法）

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本文介绍了一个算法问题，即寻找给定整数序列中满足特定条件的最长连续子序列。该子序列由三个相等长度的部分组成，首尾部分相同且中间部分与其对称。文章详细解释了解决方案，并提供了使用马拉车算法优化搜索过程的具体实现。

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Hotaru's problem

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 4237 Accepted Submission(s): 1386

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

为了方便找模板。。写个题解。。

题意：给你一个具有n(n<=1e5)个元素的整数序列，问你是否存在这样一个子序列，该子序列从左到右分为长度相等的三部分，第一部分与第三部分相同，第二部分与第一部分对称(例如 234 432 234 )，要求给出形如所述的最长连续子序列。

思路：易知234432 和432234都是回文子序列，即由前一个子序列的后半部分和后一个前半部分相同的两个回文子序列组成。

因此先用马拉车算法O(n)求出每个以 i 为中心和以 i和i+1 为中心的最长回文串长度。然后本题满足条件的只有i和i+1为中心的最长回文串长度，在纸上画画发现x>max&&p[i+x]>=x时符合条件(x=p[i]-1)，否则看 --x，直到满足条件或者小于max。最后不要忘了/2*3。

代码：

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn=2e5+10;
const int inf=0x3f3f3f3f;
int n,m,k;
int p[maxn],s[maxn],ans;
void manacher()
{
    int id=0,maxlen=0;
    for(int i=n;i>=0;i--)
    {
        s[i*2+2]=s[i];
        s[i*2+1]=-1;
    }
    s[0]=-2;
    for(int i=2;i<2*n+1;i++)
    {
        if(p[id]+id>i)p[i]=min(p[2*id-i],p[id]+id-i);
        else p[i]=1;
        while(s[i-p[i]]==s[i+p[i]])++p[i];
        if(id+p[id]<i+p[i]) id=i;
        if(maxlen<p[i]) maxlen=p[i];
    }
    //printf("%d\n",maxlen);
}
int main()
{
    int T,cas=1;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        memset(p,0,sizeof(p));
        for(int i=0;i<n;i++) scanf("%d",&s[i]);
        manacher();ans=0;
        /*for(int i=3;i<2*n+1;i+=2)
        printf(" %d",p[i]-1);*/
        for(int i=3;i<2*n;i+=2)
        if(p[i]-1>ans){
            int t=p[i]-1;
            while(t>ans&&p[i+t]<t) t--;
            ans=max(ans,t);
        }
        printf("Case #%d: %d\n",cas++,ans/2*3);
    }
    return 0;
}