Codeforces 607B Zuma (区间dp)

B. Zuma

time limit per test：2 seconds

memory limit per test：512 megabytes

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.

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Examples

input

3
1 2 1

output

input

3
1 2 3

output

input

7
1 4 4 2 3 2 1

output

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 destroy palindrome 1 2 3 2 1.

题目链接：https://codeforces.com/contest/607/problem/B

题目大意：每次可删除一段回文子串(单个字符属于回文串)，删除后剩余部分拼接在一起，求最少删几次可以将原串删空

题目分析：基础区间dp（据说是我司今年校招的笔试题之一）dp[i][j]表示区间i~j删完需要的最小次数，当s[i]==s[j]时，dp[i][j] = dp[i+1][j-1]，需特判区间长度为2的情况

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int const MAX = 505;
int n, a[MAX], dp[MAX][MAX];

int main() {
    scanf("%d", &n);
    memset(dp, 0x3f, sizeof(dp));
    for (int i = 1; i <= n; i++) {
        scanf("%d", &a[i]);
        dp[i][i] = 1;
    }
    for (int l = 1; l < n; l++) {
        for (int i = 1; i + l <= n; i++) {
            int j = i + l;
            if (a[i] == a[j]) {
                dp[i][j] = l == 1 ? 1 : dp[i + 1][j - 1];
            }
            for (int k = i; k < j; k++) {
                dp[i][j] = min(dp[i][j], dp[i][k] + dp[k + 1][j]);
            }
        }
    }
    printf("%d\n", dp[1][n]);
}

Codeforces 607B Zuma (区间dp)

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