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.
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.
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.
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
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 100060
int 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
");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
",maxx/2*3);
}
return 0;
}