dp-255C - Almost Arithmetical Progression

dp[i][j]表示最后一位是 $b_i[i]$,倒数第二位是 $b_i[j]$的最长序列长度。那么dp[i][j]=dp[j][k]+1( $b_i[i]$== $b_i[i]$)其他情况dp=2。

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cstdlib>
#include<climits>
#include<stack>
#include<vector>
#include<queue>
#include<set>
#include<bitset>
#include<map>
//#include<regex>
#include<cstdio>
#pragma GCC optimize(2)
#define up(i,a,b)  for(int i=a;i<b;i++)
#define dw(i,a,b)  for(int i=a;i>b;i--)
#define upd(i,a,b) for(int i=a;i<=b;i++)
#define dwd(i,a,b) for(int i=a;i>=b;i--)
//#define local
typedef long long ll;
typedef unsigned long long ull;
const double esp = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int inf = 1e9;
using namespace std;
ll read()
{
	char ch = getchar(); ll x = 0, f = 1;
	while (ch<'0' || ch>'9') { if (ch == '-')f = -1; ch = getchar(); }
	while (ch >= '0' && ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); }
	return x * f;
}
typedef pair<int, int> pir;
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lrt root<<1
#define rrt root<<1|1
const int N = 4010;
int dp[N][N];
int n;
int bi[N];
int main()
{
	n = read();
	upd(i, 1, n)
	{
		bi[i] = read();
	}
	upd(i, 1, n)
	{
		int k = -1;
		up(j, 1, i)
		{
			if (k == -1)dp[i][j] = 2;
			else dp[i][j] = dp[j][k] + 1;
			if (bi[j] == bi[i])k = j;
		}
	}
	int ans = 1;
	upd(i, 1, n)up(j, 1, i)ans = max(ans, dp[i][j]);
	printf("%d\n", ans);
	return 0;
}