Codeforces Round #336 (Div. 2) D. Zuma 记忆化搜索

D. Zuma

题目连接：

http://www.codeforces.com/contest/608/problem/D

Description

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.

Sample Input

3

1 2 1

Sample Output

1

Hint

题意

给你一个串，你每次可以消去一个回文串

问你最少消去多少次，可以使得这个串清空

题解：

裸的记忆化搜索，比较简单

代码

#include<bits/stdc++.h>
using namespace std;
#define maxn 805

int dp[maxn][maxn];
int vis[maxn][maxn];
int a[maxn];
int n;
int dfs(int l,int r)
{
    if(vis[l][r])return dp[l][r];
    vis[l][r]=1;dp[l][r]=1e9;
    if(l>r)return dp[l][r]=0;
    if(l==r)return dp[l][r]=1;
    if(l==r-1)
    {
        if(a[l]==a[r])return dp[l][r]=1;
        else return dp[l][r]=2;
    }
    if(a[l]==a[r])
        dp[l][r]=dfs(l+1,r-1);
    for(int i=l;i<=r;i++)
        dp[l][r]=min(dfs(l,i)+dfs(i+1,r),dp[l][r]);
    return dp[l][r];
}
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%d",&a[i]);
    cout<<dfs(1,n)<<endl;
}