CF607B: Zuma（区间dp）

最新推荐文章于 2023-03-24 15:52:51 发布

转载最新推荐文章于 2023-03-24 15:52:51 发布 · 167 阅读

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CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/junior19/p/6730087.html

本文介绍了一种解决Zuma游戏问题的方法，目标是最小化删除所有宝石所需的秒数。通过使用动态规划策略，实现了一个算法来确定最优解。

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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.

每次删除一段回文子串，把整段删除的最小次数，简单区间dp，dp[i][j] = min(dp[i][k]+dp[k+1][j], dp[i][j])，如果a[i] == a[j]，判断多一次dp[i][j] = min(dp[i+1][j-1], dp[i][j])。

# include <stdio.h>
# include <string.h>
# define INF 0x3f3f3f3f
int min(int a, int b){return a<b?a:b;}
int a[501], dp[501][501];
int main()
{
    int n, i, j, k, len;
    while(~scanf("%d",&n))
    {
        for(i=0; i<=500; ++i)
            for(j=0; j<=500; ++j)
            dp[i][j] = INF;
        for(i=0; i<n; ++i)
        {
            scanf("%d",&a[i]);
            dp[i][i] = 1;//第一次初始化
        }
        for(i=0; i<n-1; ++i)//第二次初始化
            if(a[i]==a[i+1])
                dp[i][i+1] = 1;
            else
                dp[i][i+1] = 2;
        for(len=2; len<n; ++len)
        {
            for(i=0; i+len<n; ++i)
            {
                if(a[i]==a[i+len])
                    dp[i][i+len] = dp[i+1][i+len-1];
                for(k=i; k<i+len; ++k)
                    dp[i][i+len] = min(dp[i][k]+dp[k+1][i+len], dp[i][i+len]);
            }
        }
        printf("%d\n",dp[0][n-1]);
    }
    return 0;
}

转载于:https://www.cnblogs.com/junior19/p/6730087.html