30天学会基础算法 (合唱队形—线性dp) ^7^

题目链接

题目描述

输入格式

输出格式

输入样例

8
186 186 150 200 160 130 197 220

输出样例

4

题目分析

code

#include <iostream>
#include <cstring>
using namespace std;

const int MAX = 105;
const int INF = 0x3fffffff;

int a[MAX];
int dp[MAX];
int n;

int main()
{
    cin >> n;
    for (int i = 1; i <= n; i++) cin >> a[i];
    a[0] = 0;            //初始化，空段 
    a[n + 1] = 0;

    int ans = INF;
    //枚举中间点
    for (int i = 1; i <= n; i++) 
    {
        memset(dp, 0, sizeof(dp));
        int cnt = 0;

        //左边为上升序列
        for (int j = 1; j <= i; j++) 
            for (int k = 0; k < j; k++)
                if (a[k] < a[j]) 
                    dp[j] = max(dp[j], dp[k] + 1);

        cnt += i - dp[i];        //左边出队的人数 

        //右边为下降序列
        dp[i] = 0;                //注意清零 
        for (int j = n; j >= i; j--)
            for (int k = n + 1; k > j; k--)
                if (a[k] < a[j])
                    dp[j] = max(dp[j], dp[k] + 1);
               
        cnt += n - i + 1 - dp[i];        //右边出队的人数 
        ans = min(ans, cnt);        //取最小值 
    }
    cout << ans;
    return 0;
}

• 无注释版

#include <iostream>
#include <cstring>
using namespace std;

const int MAX = 105;
const int INF = 0x3fffffff;

int a[MAX];
int dp[MAX];
int n;

int main()
{
    cin >> n;
    for (int i = 1; i <= n; i++) cin >> a[i];
    a[0] = 0; 
    a[n + 1] = 0;

    int ans = INF;
    for (int i = 1; i <= n; i++) 
    {
        memset(dp, 0, sizeof(dp));
        int cnt = 0;

        for (int j = 1; j <= i; j++) 
            for (int k = 0; k < j; k++)
                if (a[k] < a[j]) 
                    dp[j] = max(dp[j], dp[k] + 1);

        cnt += i - dp[i];

        dp[i] = 0; 
        for (int j = n; j >= i; j--)
            for (int k = n + 1; k > j; k--)
                if (a[k] < a[j])
                    dp[j] = max(dp[j], dp[k] + 1);
               
        cnt += n - i + 1 - dp[i];
        ans = min(ans, cnt); 
    }
    cout << ans;
    return 0;
}