HDU 5371 (Hotaru ' s problem-2 times palindrome string)

Hotaru ' s problemTime limit:4000/2000 MS (java/others) Memory limit:65536/65536 K (java/others)
Total submission (s): 2432 Accepted Submission (s): 841


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 than ten 9 , 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 Input

1102 3 4 4 3 2 2 3 4 4

Sample Output

Case #1:9

Authoruestc
Source2015 multi-university Training Contest 7
Recommendwange2014 | We have carefully selected several similar problems for you:5421 5420 5419 5418 5417

Match first, then violence find the second one,

Look for optimizations that are faster than the current answer.

#include <bits/stdc++.h> using namespace std; #define for (I,n) for (int. i=1;i<=n;i++) #define FORK (i,k,n) for (int i=k;i<=n;i++) #define REP (I,n) for (int. i=0;i<n;i++) #define ForD (I,n) for (int. i=n;i;i--) #define REPD (i,n) for (int i =n;i>=0;i--) #define FORP (x) for (int p=pre[x];p, p=next[p]) #define FORPITER (x) for (int &p=iter[x];p; p=next[p]) # Define Lson (x<<1) #define Rson ((x<<1) +1) #define MEM (a) memset (A,0,sizeof (a)), #define MEMI (a) memset (a,127 , sizeof (a)), #define MEMI (a) memset (A,128,sizeof (a)), #define INF (2139062143) #define F (100000007) #define MAXN (300000+ Ten) #define SP_CHAR1 ( -1) #define SP_CHAR2 ( -2) typedef long Long Ll;ll Mul (ll A,ll b) {return (a*b)%F;} ll Add (ll A,ll b) {return (a+b)%F;} ll Sub (ll A,ll b) {return (A-b+llabs (a)/f*f+f)%F; void Upd (ll &a,ll b) {a= (a%f+b%f)%F;} int A[maxn],n,r[maxn];class manacher{public:int n;int s[maxn];int p[2*maxn+2];void mem () {n=0; MEM (s) mem (p)}int str[maxn*2+2];void work () {str[0]=sp_char1; Rep (i,n) str[2*i+1]=sp_char2,Str[2*i+2]=s[i]; STR[2*N+1]=SP_CHAR2; str[2*n+2]=-3;n=2*n+2; MEM (p) int mx=0,id=0; for (i,n-1) {if (I&LT;MX) p[i]=min (P[2*id-i],mx-i), while (Str[i-p[i]]==str[i+p[i]]) ++p[i];if (Mx<i+p[i])// MX is the most right end of the identified {Mx=i+p[i];id=i;}}} S;int Main () {//freopen ("c.in", "R", stdin); int t;cin>>t; for (kcase,t) {s.mem (); scanf ("%d", &n); for (I,n) scanf ("%d", &a[i]), s.s[i-1]=a[i]; S.n=n; S.work (); int ans=0;for (int i=3;i<s.n;i+=2) {for (int j=s.p[i];j>ans;j-=2) {if (s.p[i+ (j-1)]>=j) Ans=max (ans,j  ); }}ans= (ans-1)/2*3;printf ("Case #%d:%d\n", Kcase,ans);} return 0;}

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HDU 5371 (Hotaru ' s problem-2 times palindrome string)