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python/TangDou/AcWing/LIS/482_1.cpp
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2025-08-30 18:35:01 +08:00

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#include <bits/stdc++.h>
using namespace std;
const int N = 110;
int n; // 同学的个数
int a[N]; // 山的高度数组
int f[N]; // 从左向右的最长子序列
int g[N]; // 从右向左的最长子序列
int res;
int main() {
cin >> n;
for (int i = 1; i <= n; i++) cin >> a[i];
// 正向求解 LIS问题
for (int i = 1; i <= n; i++) {
f[i] = 1;
for (int j = 1; j < i; j++)
if (a[i] > a[j]) f[i] = max(f[i], f[j] + 1);
}
// 反向求解 LIS问题
for (int i = n; i >= 1; i--) {
g[i] = 1;
for (int j = n; j > i; j--)
if (a[i] > a[j]) g[i] = max(g[i], g[j] + 1);
}
// 因为最终的那个中间点,左边计算了一次,右边双计算了一次,需要减1去重复
for (int i = 1; i <= n; i++) res = max(res, f[i] + g[i] - 1);
// 问至少要出队几人,和问“mx=满足要求的队列,最长是多长”是直接相关的,如果算得mx,则n-mx就是答案
printf("%d\n", n - res);
return 0;
}