hdu5371 Hotaru's problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2189    Accepted Submission(s): 774

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
这题可以用Manacher算法做，因为题目要找的是三段（第一段和第二段对称，第二段和第三段对称），其实就是两个连在一起的回文串，我们可以先用Manacher算法初始化各个点的p[i]值（即可以向右延伸的最大距离，包括本身，这时已经加入了-1代替算法中的'#',-2代替算法中的'$'），然后对于每个i，枚举j（j属于1~p[i]-1),如果i+j-p[i+j]+1<=i，那么说明i，j可以分别作为第一、二段的点和第二、三段的点）。
这里有个优化，因为枚举时满足条件的只有'#'（即'-1’），所以我们可以使i，j每次变化2.
#include<iostream>#include<stdio.h>#include<stdlib.h>#include<string.h>#include<math.h>#include<vector>#include<map>#include<set>#include<queue>#include<stack>#include<string>#include<algorithm>using namespace std;#define maxn 100060int a[maxn],b[2*maxn],p[2*maxn];int main(){    int n,m,i,j,T,mx,idx,maxx,num1=0;    scanf("%d",&T);    while(T--)    {        scanf("%d",&n);        for(i=0;i<n;i++){            scanf("%d",&a[i]);        }        if(n<3){            printf("0\n");continue;        }        b[0]=-2;        b[1]=-1;        for(i=0;i<n;i++){            b[i*2+2]=a[i];            b[i*2+3]=-1;        }        n=2*n+2;mx=0;        for(i=0;i<n;i++){            if(i<mx){                p[i]=min(p[idx*2-i],mx-i);            }            else p[i]=1;            while(b[i-p[i]]==b[i+p[i]]){                p[i]++;            }            if(mx<i+p[i]){                mx=i+p[i];                idx=i;            }        }        maxx=0;        for(i=3;i<n;i+=2){            for(j=p[i]-1;j>=1;j-=2){                if(j<maxx)break;                if(i+j-p[i+j]+1<=i){                    maxx=max(maxx,j);break;                }            }        }        num1++;        printf("Case #%d: ",num1);        printf("%d\n",maxx/2*3);    }    return 0;}