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.

The first line of input contains a single integer n

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.

Print a single integer — the minimum number of seconds needed to destroy the entire line.

3
1 2 1

1

3
1 2 3

3

7
1 4 4 2 3 2 1

2

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.

題意：給你一個序列，每次可以刪除一個回文串，問最少可以幾次全部刪完。

題解：用f[i][j]表示將[i,j]全部刪完的最小次數，對於每一個a[i]==a[j]，f[i][j]就一定可以由f[i+1][j-1]轉移過來，因為f[i+1][j-1]最後一步一定是一個回文串。再枚舉分界點k,f[i][j]<-f[i][k]+f[k+1][j];

代碼：

#include<cstdio>
#include<cstring>
#include<algorithm>
#define inf 2000000000
using namespace std;
template<class T>inline void read(T &x)
{
    x=0;char ch=getchar();
    while(ch<‘0‘||ch>‘9‘)ch=getchar();
    while(ch<=‘9‘&&ch>=‘0‘)x=(x<<1)+(x<<3)+(ch^48),ch=getchar();
    return;
}
int f[510][510],a[510];
int main()
{
    int n;
    read(n);
    memset(f,0x7f,sizeof(f));
    for(int i=1;i<=n;i++)read(a[i]);
    for(int i=1;i<=n;i++)f[i][i]=1;
    for(int l=2;l<=n;l++)
    for(int i=1;i+l-1<=n;i++)
    {
        int j=i+l-1;
        if(l==2&&a[i]==a[j])f[i][j]=1;
        else if(a[i]==a[j]&&l>2)f[i][j]=f[i+1][j-1];
        for(int k=i;k<j;k++)
        f[i][j]=min(f[i][j],f[i][k]+f[k+1][j]);
    }
    printf("%d\n",f[1][n]);
    return 0;
}

Codeforces 607BZuma(區間dp)