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 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

2 3 4 4 3 2 2 3 4 4

Sample Output

Case #1: 9

题目大意：给你t组数据，一个n，下面一行有n个数，问你能形成形如abccbaabc这样的序列长度最长是多少。

解题思路：先用manacher处理出来串的所有字符的回文半径。然后枚举第一段跟第二段回文位置i，从i+p[i]-->i进行枚举第二段跟第三段回文位置j。如果以j为回文中心的左端能小于等于i的位置，说明满足要求，更新结果。

#include<bits/stdc++.h>

using namespace std;

#define min(a,b) ((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++){    //需要处理到n+1

        if(i<mx){

            p[i]=min(p[id*2-i],mx-i);   //这里写的时候写成mx-id，SB了。

        }else{

            p[i]=1;

        }

        for(;a[i+p[i]]==a[i-p[i]];++p[i]);  //-2，-3防越界

        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){ //逆序枚举。manacher算法保证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;

}

巴特西

HDU 5371——Hotaru's problem——————【manacher处理回文】

Hotaru's problem

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