2020牛客暑期多校训练营（第五场）D.Drop Voicing

2020牛客暑期多校训练营（第五场）D.Drop Voicing

题目链接

题目描述

Inaka composes music. Today’s arrangement includes a chord of nn notes that are pairwise distinct, represented by a permutation $p_{1\dots n}$ of integers from 1 to n (inclusive) denoting the notes from the lowest to the highest.

Her friend, Miyako, sneaks in and plays a trick by altering the chord with the following two operations:

• Drop-2: Take out the second highest note and move it to the lowest position, i.e. change the permutation to $p_{n-1},p_1,p_2,\dots ,p_{n-3},p_{n-2},p_n$.
• Invert: Take out the lowest note and move it to the highest position, i.e. change the permutation to $p_2,p_3,\dots ,p_{n-1},p_n,p_1$.

Any number of consecutive Drop-2 operations is considered a multi-drop. Miyako would like to change the permutation to an ordered permutation, $1,2,\dots ,n$, in the fewest number of multi-drops possible. Please help her find the number of multi-drops needed.

输入描述:

The first line contains an integer $n(2\le n\le 500)$ — the number of notes.

The second line contains n space-separated integers $p_1,p_2,\dots ,p_n$ — the original permutation of notes.

The input guarantees each integer from 1 to n (inclusive) appears in the permutation exactly once.

输出描述:

Output one integer — the number of multi-drops required to change the permutation to an ordered one.

示例1

输入

6
2 4 5 1 3 6

输出

2

示例2

输入

8
8 4 7 3 6 2 5 1

输出

5

这题其实是非常简单的 $dp$，很多人容易被题目要求迷惑，我们试想，要求一个序列有序，很容易想到最长上升子序列，保持最长的子序列，调整其他位置即可，而我们看向 $invert$ 操作，很明显提示你将序列看作一个环，而且细心一点就会发现，用 $invert$ 搭配 $drop$ 操作就可以将任意数字移到任意位置，那么题目到这里就很明确了，无非就是求一个环的最长上升子序列，剩下的元素个数就是需要操作的次数了，还有一个小细节，注意看题目数据大小，数据小一般就提示 DP 了，AC代码如下：

#include
using namespace std;
typedef long long ll;
const int N=505;
int n,a[N],dp[N];
int main(){
    int n,ans=0;
    scanf("%d",&n);
    for(int i=0;i<n;i++) scanf("%d",&a[i]);
    for(int i=0;i<n;i++){
        fill(dp,dp+n,1);
        for(int j=0;j<n;j++){
            for(int k=0;k<j;k++){
                if(a[j]>a[k]) dp[j]=max(dp[j],dp[k]+1);
            }
        }
        ans=max(ans,dp[n-1]);
        rotate(a,a+1,a+n);
    }
    printf("%d",n-ans);
	return 0;
}