HDU 4513 Series Story--Perfect formation ii_ string

A series of stories about Jill--Perfect Formation II time limit:3000/1000 MS (java/others) Memory limit:65535/32768 K (java/others)
Total submission (s): 4024 accepted Submission (s): 1602


Problem Description Gigo again came up with a new perfect formation game.
Suppose there are n individuals standing in front of him in order, their height is h[1], h[2] ... h[n], Gigo hope to pick out some people, let these people form a new formation, if the new formation meet the following three requirements, it is the new perfect formation:

1, the selected person to maintain the original formation of the relative order unchanged, and must be in the original formation of continuous;
2, left-right symmetry, with the formation of M-personal form of new formations, then the 1th person and the height of the same m, the 2nd person and the m-1 personal height of the same, and so on, of course, if M is odd, the middle of the person can be arbitrary;
3, from the left to the middle of that person, height should be guaranteed not to decline, if using H to indicate the height of the new formation, then h[1] <= h[2] <= h[3] ... <= h[mid].

Now Gigo wants to know how many people can be chosen to form a new perfect formation.
Input input data The first row contains an integer t, which indicates a total of T-Group test data (T <= 20);
Each group of data first is an integer n (1 <= n <= 100000), indicating the number of the original formation, the next line input n integers, indicating the original formation from left to right station of the height of people (<= H <= 250, do not exclude particularly short and tall).
Output can be composed of the largest number of perfect formation, each set of output occupies one row.
Sample Input

2 3 51 52 51 4 51 52 52 51
Sample Output

3 4
Source 2013 Tencent Programming Marathon second (March 22) can be seen from the request is a palindrome a string of numbers, can still use the manacher algorithm, in the calculation of the P[i] array, if the new array is not not a position of the number (2), Then we have to judge it with the front.

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int MAXN=1E5+10;
int NUM[MAXN];
int num_new[maxn<<1];
int p[maxn<<1];
int n;
int init () {
	num_new[0]=-1;
	num_new[1]=-2;
	int j=2;
	for (int i=0;i<n;i++) {
		num_new[j++]=num[i];
		num_new[j++]=-2;
	}
	num_new[j]=-2;
	return j;
}
int Manacher () {
	int len=init ();
	int id;
	int maxx=0;
	int max_len=-1;
	for (int i=1;i<len;i++) {
		if (I<maxx) p[i]=min (p[2*id-i],maxx-i);
		else p[i]=1;
		while (Num_new[i+p[i]]==num_new[i-p[i]]) {
			if (num_new[i+p[i]]!=-2) {
				if (num_new[i+p[i]]<=num_new[i+p [i]-2]) p[i]++;
				else break;
			}
			p[i]++;
		}
		if (Maxx<i+p[i]) {
			maxx=i+p[i];
			id=i;
		}
		Max_len=max (max_len,p[i]-1);
	}
	return max_len;
}
int main () {
	int test;
	scanf ("%d", &test);
	while (test--) {
		scanf ("%d", &n);
		for (int i=0;i<n;i++) {
			scanf ("%d", &num[i]);
		printf ("%d\n", Manacher ());
	}
	return 0;
}