思路:用两个最长上升子序列结合起来求解
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 110;
int n;
int h[N]; //h存数列,f和g存状态
int f[N], g[N];
int main()
{
cin >> n;
for (int i = 1; i <= n; i ++ ) cin >> h[i];
for (int i = 1; i <= n; i ++ ) //从左往右找最长上升子序列
{ //最长上升子序列为DP模板题,背就完了
f[i] = 1;
for (int j = 1; j < i; j ++ )
if (h[j] < h[i])
f[i] = max(f[i], f[j] + 1);
}
for (int i = n; i; i -- ) //从右往左找最长下降子序列
{
g[i] = 1;
for (int j = n; j > i; j -- )
if (h[j] < h[i])
g[i] = max(g[i], g[j] + 1);
}
int res = 0;
for (int k = 1; k <= n; k ++ )
res = max(res, f[k] + g[k] - 1); //结合以上两种最长子序列找出最长合唱队形
//俩序列的值都是包括了当前小朋友的,所以减去1
cout << n - res << endl;
return 0;
}