HDU5371 Hotaru's problem

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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

 

 

正解：manacher

解题报告：

　　求形如ABA（B与A对称0）的串的最大长度。

　　先用manacher预处理出以每个点为中心的最长回文子串长度，枚举第二个部分即B的起点，再枚举串的长度，发现可行则更新答案（代码中非常清楚了）。可以用已经得到的ans剪枝。

　　辣鸡HDU评测速度飘忽不定，刷了几次才跑过。

 

 

//It is made by ljh2000
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;
typedef long long LL;
const int MAXN = 200011;
int n,len,a[MAXN],p[MAXN],ans;
inline int getint(){
    int w=0,q=0; char c=getchar(); while((c<'0'||c>'9') && c!='-') c=getchar();
    if(c=='-') q=1,c=getchar(); while (c>='0'&&c<='9') w=w*10+c-'0',c=getchar(); return q?-w:w;
}

inline void manacher(){
	//memset(p,0,sizeof(p));
	for(int i=1;i<=n;i++) p[i]=0;
	int maxR=0,id=0;
	for(int i=1;i<=n;i++) {
		if(i<maxR) p[i]=min(p[2*id-i],maxR-i);
		else p[i]=1;
		for(;i+p[i]<=n && a[i-p[i]]==a[i+p[i]];p[i]++) ;
		if(i+p[i]>maxR) { maxR=i+p[i]; id=i; }
	}
}

inline void work(){
	int T=getint(); int Case=0;
	while(T--) {
		len=getint(); a[0]=-2; a[1]=-1; n=1; for(int i=1;i<=len;i++) { a[++n]=getint(); a[++n]=-1; }
		manacher(); ans=1;
		//+2的原因是对称中心只能取在区分符号上
		for(int i=3;i<n/*!!!*/;i+=2) //枚举the second part
			for(int j=ans;j<=p[i];j+=2) //枚举一个部分的长度，利用ans剪枝
				if(p[i+j-1]>=j) //如果the third part的回文长度大于等于j，即j可以作为答案（part2和part3对称）
					ans=j;
		Case++; ans/=2; ans*=3;//只算了两个part的长度
		printf("Case #%d: %d\n",Case,ans);
	}
}

int main()
{
    work();
    return 0;
}