题意

题目链接

Sol

这题最直接的维护区间以0/1结尾的LIS的方法就不说了。

其实我们可以直接考虑翻转以某个位置为中点的区间的最大值

不难发现前缀和后缀产生的贡献都是独立的，可以直接算。维护一下前缀/后缀和即可

#include<bits/stdc++.h>

#define Pair pair<int, int>

#define MP(x, y) make_pair(x, y)

#define fi first

#define se second

#define LL long long

#define ull unsigned long long

#define Fin(x) {freopen(#x".in","r",stdin);}

#define Fout(x) {freopen(#x".out","w",stdout);}

using namespace std;

const int MAXN = 501, mod = 1e9 + 7, INF = 1e9 + 10;

const double eps = 1e-9;

template <typename A, typename B> inline bool chmin(A &a, B b){if(a > b) {a = b; return 1;} return 0;}

template <typename A, typename B> inline bool chmax(A &a, B b){if(a < b) {a = b; return 1;} return 0;}

template <typename A, typename B> inline LL add(A x, B y) {if(x + y < 0) return x + y + mod; return x + y >= mod ? x + y - mod : x + y;}

template <typename A, typename B> inline void add2(A &x, B y) {if(x + y < 0) x = x + y + mod; else x = (x + y >= mod ? x + y - mod : x + y);}

template <typename A, typename B> inline LL mul(A x, B y) {return 1ll * x * y % mod;}

template <typename A, typename B> inline void mul2(A &x, B y) {x = (1ll * x * y % mod + mod) % mod;}

template <typename A> inline void debug(A a){cout << a << '\n';}

template <typename A> inline LL sqr(A x){return 1ll * x * x;}

inline int read() {

    char c = getchar(); int x = 0, f = 1;

    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}

    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();

    return x * f;

}

int N, a[MAXN], f[MAXN], g[MAXN];

signed main() {

	N = read();

	for(int i = 1; i <= N; i++) a[i] = read(), f[i] = f[i - 1] + (a[i] == 1);

	for(int i = N; i >= 1; i--) g[i] = g[i + 1] + (a[i] == 2);

	int ans = 0;

	for(int i = 1; i <= N; i++) {

		int s1 = 0, s2 = 0;

		for(int j = i; j >= 1; j--) chmax(s1, f[j - 1] + g[j] - g[i]);

		for(int j = i; j <= N; j++) chmax(s2, g[j + 1] + f[j] - f[i - 1]);

		chmax(ans, max(s1 + s2, f[i - 1] + g[i]));

	}

	cout << ans;

    return 0;

}