HDOJ5371 manacher简单运用

Hotaru's problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 3878    Accepted Submission(s): 1271

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
  
  

 

Author

UESTC

 

Source

2015 Multi-University Training Contest 7

 

Recommend

wange2014   |   We have carefully selected several similar problems for you:   6193  6192  6191  6190  6189 

两个回文串之间有重叠的部分而已，跑一遍manacher后枚举即可。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
using namespace std;

const int maxn = 1e5+50;
int i,j,k,n,id,ans,cnt;
int s[maxn<<1],a[maxn<<1];

void init(){
     scanf("%d",&n);
     s[0] = -1; //这里一定要跟其他新添加的非原串字符相同，不然样例都过不了。。。
     cnt = 0;
     for (i=0; i<n; i++) {scanf("%d",&s[++cnt]); s[++cnt]= -1;}
     cnt++;
}

void manacher(){
     ans = 0; id = 0;
     memset(a,0,sizeof(a));
     int mx = 0;
     for (i=0; i<cnt; i++) {
        if (mx>i) a[i] = min(a[id*2-i],mx-i);
        else a[i] = 1;
        while (s[i-a[i]]==s[i+a[i]]) a[i]++;
        if (mx<i+a[i]) {id = i; mx = a[i] + i;}
     }

     for (i=0; i<cnt; i+=2)
        for (j=ans; j<a[i]; j+=2){
         if (a[i+j]>j) ans = j;
     }
}

int main(){
    int T;
    cin >> T;
    for (int Case=1; Case<=T; Case++) {
          init();
          manacher();
          printf("Case #%d: %d\n",Case,ans*3/2);
    }
    return 0;
}