HDU 5371--hotaru ' s problem —————— "Manacher treatment palindrome"

Hotaru ' s problem

Time limit:4000/2000 MS (java/others) Memory limit:65536/65536 K (java/others)
Total submission (s): 1765 Accepted Submission (s): 635

Problem Descriptionhotaru Ichijou recently is addicated to math problems. Now she's 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 of the same as the thrid part,
2. The first part and the second part is symmetrical.
For example, the sequence 2,3,4,4,3,2,2,3,4 are a n-sequence, which the first part 2,3,4 are the same as the Thrid Part 2, 3, 4, the first part 2,3,4 and the second part 4,3,2 is symmetrical.

Give you n positive intergers, your task was to find the largest continuous sub-sequence, which is n-sequence.

Inputthere is 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 than9 , descripting a sequence.

Outputeach case contains only one line. Should start with ' case #i: ', with I implying the case number, followed by a integer, the largest length of n-se Quence.

We guarantee the sum of all answers are less than 800000.

Sample Input1102 3 4 4 3 2 2 3 4 4

Sample outputcase #1:9  topic: give you T Group of data, an n, the following line has n number, ask you can form like abccbaabc such a sequence length of the longest is how much.   Problem Solving ideas: first with manacher processing out the string of all the characters of the palindrome radius. Then enumerate the first paragraph and the second section of the palindrome position I, from I+p[i]-->i to enumerate the second paragraph and the third section palindrome position J. If J is the center of the left end can be less than equal to the position of I, to meet the requirements, update the results.  

#include <bits/stdc++.h>using namespace std; #define MIN (a) < (b) (    A):(B)) const int Maxn=1e6;int a[maxn],p[maxn];void manacher (int n) {a[0]=-2;a[n+1]=-1;a[n+2]=-3;    int mx=0,id=0;   for (int i=1;i<=n+1;i++) {//needs to be processed to n+1 if (i<mx) {p[i]=min (p[id*2-i],mx-i);        It was written in MX-ID,SB.        }else{p[i]=1;  } for (; a[i+p[i]]==a[i-p[i]];++p[i]);            -2,-3 Cross-border if (i+p[i]>mx) {mx=p[i]+i;        Id=i;    }} for (int i=1;i<=n+1;++i) {--p[i];    }}int Main () {//Freopen ("1003.in", "R", stdin);//Freopen ("OUTTTT.txt", "w", stdout);    int t,n,cnt=0;    scanf ("%d", &t);        while (t--) {scanf ("%d", &n);            for (int i=1;i<=2*n;i+=2) {a[i]=-1;        scanf ("%d", &a[i+1]);        } manacher (2*n);        int maxv=0,j; for (int i=1;i<=2*n;i+=2) {//2*n for (j=i+p[i];j-maxv>i;j-=2) {//Reverse enumeration.              Manacher algorithm guarantees j<=2*n+1   if (j-p[j]<=i) {maxv=j-i;                Break        }}} maxv=3* (MAXV/2);    printf ("Case #%d:%d\n", ++CNT,MAXV); } return 0;}

　　

HDU 5371--hotaru ' s problem —————— "Manacher treatment palindrome"