C. Tenzing and Balls

链接：https://codeforces.com/problemset/problem/1842/C or https://www.luogu.com.cn/problem/CF1842C

大概的思路就是利用dp[i]记录前i个数据最多消掉的数字个数，然后对∀j：a[i] == a[j] && j < i进行dp[i] = dp[j-1] + i-j+1的递推
优化：

代码：

#define _CRT_SECURE_NO_WARNINGS 
#include<iostream>
#include<vector>
#include<algorithm>
#include<math.h>
#include<sstream>
#include<string>
#include<string.h>
#include<iomanip>
#include<stdlib.h>
#include<map>
#include<queue>
#include<limits.h>
#include<climits>
#include<fstream>
#include<stack>
typedef long long ll;
using namespace std;

const int N = 2e5 + 10;
ll dp[N],a[N],mp[N];
int main()
{
	ios::sync_with_stdio(false); cin.tie(0), cout.tie(0);
	ll t; cin >> t;
	while (t--)
	{
		ll n; cin >> n;
		memset(dp, 0, sizeof(dp));
		memset(mp, -0x3f3f3f3f, sizeof(mp));//这个初始设置成最小
		for (ll i = 1; i <= n; i++)
		{
			ll x; cin >> x;
			a[i] = x;
		}
		dp[1] = 0;//dp:消掉的数量
		for (ll i = 1; i <= n; i++)
		{
			//dp[i] =  max(dp[j - 1] + i - j + 1,dp[i]);
			dp[i] = max(dp[i - 1], mp[a[i]] + i);
			mp[a[i]] = max(mp[a[i]], dp[i - 1] - i + 1);
		}
		cout << dp[n]<<'\n';

	}
	return 0;
}