Codeforces Round #336 (Div. 2) D. Zuma 区间dp

D. Zuma

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 test(s)

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.

题意：给你一个串，每次可以消除 其一个回文子串，问你最少消除几次

题解：设dp[l][r] 为 l到r 需要消除几次  ，

搜索就好

//meek///#include<bits/stdc++.h>
#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <queue>
#include <map>
#include <set>
#include <stack>
#include <sstream>
#include <vector>
using namespace std ;
#define mem(a) memset(a,0,sizeof(a))
#define pb push_back
#define fi first
#define se second
#define MP make_pair
typedef long long ll;

const int N = ;
const int inf = ;
const int mod= ;

int dp[N][N],a[N],n;
int dfs(int l,int r) {
    if(dp[l][r] != -) return dp[l][r];
   int& ret=dp[l][r]=inf;
   if(l==r) ret=;
   else if(r-l == ) {
    if(a[l]==a[r]) ret = ;
    else  ret  = ;
   }
   else {
        if(a[l] == a[r]) ret = dfs(l+,r-);
    for(int i=l;i<r;i++) ret = min(ret,dfs(l,i)+dfs(i+,r));
   }
    return ret;
}
int main() {
    scanf("%d",&n);
    memset(dp,-,sizeof(dp));
    for(int i=;i<=n;i++) {
        scanf("%d",&a[i]);
    }
    printf("%d\n",dfs(,n));
    return ;
}

代码