Hotaru‘s problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

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:  5421 5420 5419 5418 5417

先匹配第一个，再暴力找第二个，

注意只找比当前答案优的优化速度

#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+10)
#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-b)/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<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为已查明的最右端
			{
				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;
}

版权声明：本文为博主原创文章，未经博主允许不得转载。

HDU 5371(Hotaru's problem-2次回文串)