HDU-5371 Manacher+线段树维护

Hotaru's problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 3123    Accepted Submission(s): 1057

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

1102 3 4 4 3 2 2 3 4 4

 

Sample Output

Case #1: 9

 

Author

UESTC

 

Source

2015 Multi-University Training Contest 7

题目链接：

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

题目大意：

定义一种N-sequence，由三部分组成，第一部分与第二部分对称，第一部分与第三部分相同。给出一个n个数的序列，问其中最长的N-sequence的长度。

解题思路：

Manacher预处理，若有以i为中心的回文串s1和以j为中心的回文串s2，且s1的右半部分覆盖了j，s2的左半部分覆盖了i，那么可形成一个长度为(j-i)/2*3的N-sequence。从左到右枚举i，每次查询[i-pp[i]+1,i]中到i最远的被标记的中心位置，然后将位置i标记，到位置i+pp[i]-1处再将标记i去掉，查询与更新标记用线段树实现即可。

AC代码：

import java.io.*;import java.util.*;public class Main {static StreamTokenizer in=new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));static int nextInt() throws IOException    {      in.nextToken();        return (int)in.nval;     }static PrintWriter out=new PrintWriter(new OutputStreamWriter(System.out));static int T,n,len,inf=(int)1e9+7;static int ans,res,ee;static int[] aa=new int[100005];static int[] ss=new int[200005];static int[] pp=new int[200005];static int[] min=new int[800005];static int[] head=new int[200005];static int[] next=new int[200005];static int[] edge=new int[200005];static void add(int a,int b){edge[ee]=b;next[ee]=head[a];head[a]=ee++;}static void updata(int a,int k,int l,int r,int x){if(l==r) { min[k]=x;return;}int mid=(l+r)/2;if(a<=mid) updata(a,k*2+1,l,mid,x);else updata(a,k*2+2,mid+1,r,x);min[k]=Math.min(min[k*2+1],min[k*2+2]);}static int query(int a,int b,int k,int l,int r){if(a>r||b<l) return inf;if(a<=l&&r<=b) return min[k];int mid=(l+r)/2;int vl=query(a,b,k*2+1,l,mid);int vr=query(a,b,k*2+2,mid+1,r);return Math.min(vl,vr);}static void Manacher(){Arrays.fill(ss,-1);for(int i=1;i<=n;i++)ss[i*2]=aa[i];ss[0]=-2;len=n*2+1;int max=0,id=0;for(int i=1;i<=len;i++){if(max>i)pp[i]=Math.min(max-i,pp[id*2-i]);else pp[i]=1;while(ss[i+pp[i]]==ss[i-pp[i]])pp[i]++;if(i+pp[i]>max) { max=i+pp[i];id=i;}}}public static void main(String[] args) throws IOException {//Scanner in=new Scanner(System.in);T=nextInt();for(int t=1;t<=T;t++){n=nextInt();for(int i=1;i<=n;i++)aa[i]=nextInt();Manacher();ans=0;Arrays.fill(min,inf);Arrays.fill(head,-1);ee=0;for(int i=1;i<=len;i+=2){res=query(i-pp[i]+1,i,0,1,len);ans=Math.max(ans,i-res+1);updata(i,0,1,len,i);add(i+pp[i]-1,i);for(int j=head[i];j!=-1;j=next[j])updata(edge[j],0,1,len,inf);}ans=ans/2*3;out.println("Case #"+t+": "+ans);}out.flush();}}