ABC 397

C. Variety Split Easy

        对于一个数组在某处切开分成两段的问题，处理方法是维护一个前缀信息和一个后缀信息

 

#include<bits/stdc++.h>
#define int long long
using namespace std;
const int N = 3e5 + 5, INF = 1e18;

int T, n, cnt, ans, f[N], g[N], a[N];
map<int, int> mp1, mp2;

signed main()
{
	cin >> n;
	for (int i = 1; i <= n; i ++)
		cin >> a[i];
	for (int i = 1; i <= n; i ++)
		if (mp1.count(a[i]) == 0)
		{
			f[i] = f[i - 1] + 1;
			mp1[a[i]] = 1;
		}
		else
			f[i] = f[i - 1];
	for (int i = n; i >= 1; i --)
		if (mp2.count(a[i]) == 0)
		{
			g[i] = g[i + 1] + 1;
			mp2[a[i]] = 1;
		}
		else
			g[i] = g[i + 1];
	for (int i = 1; i < n; i ++)
		ans = max(ans, f[i] + g[i + 1]);
	cout << ans;
	return 0;
}