CodeForces - 607B Zuma 구간 dp

Genos recently installed the game Zuma on his phone. In Zuma there exists a line of n gemstones, the i-th of which has color ci. The goal of the game is to destroy all the gemstones in the line as quickly as possible.
In one second, Genos is able to choose exactly one continuous substring of colored gemstones that is a palindrome and remove it from the line. After the substring is removed, the remaining gemstones shift to form a solid line again. What is the minimum number of seconds needed to destroy the entire line?
Let us remind, that the string (or substring) is called palindrome, if it reads same backwards or forward. In our case this means the color of the first gemstone is equal to the color of the last one, the color of the second gemstone is equal to the color of the next to last and so on.
Input
The first line of input contains a single integer n (1 ≤ n ≤ 500) — the number of gemstones.
The second line contains n space-separated integers, the i-th of which is ci (1 ≤ ci ≤ n) — the color of the i-th gemstone in a line.
Output
Print a single integer — the minimum number of seconds needed to destroy the entire line.
Examples
Input

3
1 2 1

Output

1

Input

3
1 2 3

Output

3

Input

7
1 4 4 2 3 2 1

Output

2

Note
In the first sample, Genos can destroy the entire line in one second.
In the second sample, Genos can only destroy one gemstone at a time, so destroying three gemstones takes three seconds.
In the third sample, to achieve the optimal time of two seconds, destroy palindrome 4 4 first and then destroy palindrome 1 2 3 2 1.
제목: 제목 뜻: 서열을 하나 줍니다. 매번 답장 서열을 제거할 수 있습니다. 최소한 몇 번의 조작이 필요하면 이 서열을 삭제할 수 있는지 물어보십시오.
문제풀이: 본보기인 셈이다
dp[l][r] = min(dp[l][r], dp[l][k] + dp[k + 1][r]);  if(a[l] == a[k]) dp[l][r] = min(dp[l][r], dp[l + 1][k - 1] + dp[k + 1][r] + (l + 1 > k - 1));//만약 l에서 k 사이가 또 있다면 병합하고, l에서 k가 세지 않으면 한 번이라도

#include 
using namespace std;

const int N = 510;

int n, m;
int a[N], dp[N][N];

int main() {
	
	while(~scanf("%d", &n)) {
		memset(dp, 0, sizeof(dp));
		for(int i = 1; i <= n; i++) {
			scanf("%d", &a[i]);
			dp[i][i] = 1;
		}

		for(int len = 2; len <= n; len++) {
			for(int l = 1; l <= n - len + 1; l++) {
				int r = l + len - 1;
				dp[l][r]  = len;
				for(int k = l; k <= r; k++) {
					dp[l][r] = min(dp[l][r], dp[l][k] + dp[k + 1][r]);
					if(a[l] == a[k]) dp[l][r] = min(dp[l][r], dp[l + 1][k - 1] + dp[k + 1][r] + (l + 1 > k - 1));
 				}
			}
		}
		printf("%d
", dp[1][n]);
	}
	return 0;
}