Hdu4512: gigo series of stories-perfect formation I (LICs)

Problem description gigo has been interested in the formation over the past few days.
One day, N people stood in front of him in sequence. Their heights were H [1], H [2]... H [N], gigo hopes to pick out some people from them to form a new formation. If the new formation meets the following three requirements, it is called a perfect formation:
　　
1. The picked persons remain unchanged in the relative order of the original formation;
2. symmetric between the left and right. If m people form a new formation, the height of 1st people is the same as that of m people, and that of 2nd people is the same as that of m people, if M is an odd number, the person in the middle can be any;
3. The person from left to middle must increase his height. If H is used to represent 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 selected to form a perfect formation?

 

Input T in the first line indicates a total of T groups of data (T <= 20 );
Input N (1 <= n <= 200) for each data group, and then input n integers in the next row, the height of the person standing in the original formation from left to right (50 <= H <= 250, especially short and tall ).

Output please output the maximum number of people that can form a perfect formation, and each group of data output occupies one row.

Sample Input

2351 52 51451 52 52 51

 

Idea: it is easy to think of turning the original string and then finding LICs,

# Include <stdio. h> # include <string. h ># include <algorithm> using namespace STD; int n, a [205], B [205], DP [205]; int LICs () {int Max, I, j, k; memset (DP, 0, sizeof (DP); max = 0; for (I = 1; I <= N; I ++) {k = 0; for (j = 1; j <= n-I + 1; j ++) {if (a [I] = B [J]) {If (J! = (N-I + 1) // does not match itself {If (DP [J] <(DP [k] + 2 )) // Add 2 DP [J] = DP [k] + 2 ;} else // matches itself {If (DP [J] <(DP [k] + 1 )) // Add only 1 DP [J] = DP [k] + 1 ;}} else if (a [I]> B [J] & DP [k] <DP [J]) k = J; If (max <DP [J]) max = DP [J] ;}} return Max ;}int main () {int T, I; scanf ("% d", & T); While (t --) {scanf ("% d", & N); for (I = 1; I <= N; I ++) scanf ("% d ", & A [I]); for (I = 1; I <= N; I ++) B [I] = A [n-I + 1]; printf ("% d \ n", LICs ();} return 0 ;}