【44.19%】【codeforces 608D】Zuma

time limit per test2 seconds

memory limit per test512 megabytes

inputstandard input

outputstandard output

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.

【题目链接】:http://codeforces.com/contest/608/problem/D

【题解】



记忆化搜索;

jyh(l,r);表示把l..r这个区间范围的数字全部去掉最少需要的操作次数;

如果a[l]==a[r];则整个问题转化为为jyh(l+1,r-1);

当然如果a[l]!=a[r]则还没完;

需要枚举一下分割点i;

把整个问题转化为l..i,和i+1..r两个部分;

如果i==l,则说明最左边那个单独一个取出来操作一次(自己单独一个肯定是回文);

如果l==r返回1;

如果l==r-1->(且a[l]==a[r]返回1，如果a[l]!=a[r]返回2;)

掌握一下那个分割序列的方法;

其他的则没有什么了;

最多就500*500个状态;



【完整代码】

#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <set>
#include <map>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <queue>
#include <vector>
#include <stack>
#include <string>
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long

using namespace std;

const int MAXN = 510;
const int INF = 0x3f3f3f3f;
const int dx[5] = {0,1,-1,0,0};
const int dy[5] = {0,0,0,-1,1};
const double pi = acos(-1.0);

int n,a[MAXN],f[MAXN][MAXN];

void rel(LL &r)
{
    r = 0;
    char t = getchar();
    while (!isdigit(t) && t!='-') t = getchar();
    LL sign = 1;
    if (t == '-')sign = -1;
    while (!isdigit(t)) t = getchar();
    while (isdigit(t)) r = r * 10 + t - '0', t = getchar();
    r = r*sign;
}

void rei(int &r)
{
    r = 0;
    char t = getchar();
    while (!isdigit(t)&&t!='-') t = getchar();
    int sign = 1;
    if (t == '-')sign = -1;
    while (!isdigit(t)) t = getchar();
    while (isdigit(t)) r = r * 10 + t - '0', t = getchar();
    r = r*sign;
}

int jyh(int l,int r)
{
    if (f[l][r]!=INF)
        return f[l][r];
    int *k = &f[l][r];
    if (l == r)
        return f[l][r] = 1;
    if (l==r-1)
    {
        if (a[l]==a[r])
            return f[l][r]=1;
        else
            return f[l][r] = 2;
    }
    if (a[l] == a[r])
        *k = jyh(l+1,r-1);
    for (int i = l;i <= r;i++)
        *k = min(*k,jyh(l,i)+jyh(i+1,r));
    return f[l][r];
}

int main()
{
   // freopen("F:\\rush.txt","r",stdin);
    memset(f,INF,sizeof(f));
    rei(n);
    for (int i = 1;i <= n;i++)
        rei(a[i]);
    printf("%d\n",jyh(1,n));
    return 0;
}