cf255C Almost Arithmetical Progression

最新推荐文章于 2017-07-19 18:21:37 发布

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原文链接：http://www.cnblogs.com/scau-zk/p/5654397.html

本文介绍了一种算法，用于解决寻找最长几乎等差子序列的问题。该问题定义了一个几乎等差序列，并通过动态规划的方法高效求解最长子序列。文章详细解释了算法思路并给出了实现代码。

C. Almost Arithmetical Progression

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

Gena loves sequences of numbers. Recently, he has discovered a new type of sequences which he called an almost arithmetical progression. A sequence is an almost arithmetical progression, if its elements can be represented as:

a1 = p, where p is some integer;
ai = ai - 1 + ( - 1)i + 1·q (i > 1), where q is some integer.

Right now Gena has a piece of paper with sequence b, consisting of n integers. Help Gena, find there the longest subsequence of integers that is an almost arithmetical progression.

Sequence s1, s2, ..., sk is a subsequence of sequence b1, b2, ..., bn, if there is such increasing sequence of indexes i1, i2, ..., ik (1 ≤ i1 < i2 < ... < ik ≤ n), that bij = sj. In other words, sequence s can be obtained from b by crossing out some elements.

Input

The first line contains integer n (1 ≤ n ≤ 4000). The next line contains n integers b1, b2, ..., bn (1 ≤ bi ≤ 106).

Output

Print a single integer — the length of the required longest subsequence.
Examples
input
2
3 5
output
2
input
4
10 20 10 30
output
3
Note

In the first test the sequence actually is the suitable subsequence.

In the second test the following subsequence fits: 10, 20, 10.

dp[i][j]表示以a[i]为尾，a[j]是这个序列前一个数的最长长度

dp[i][j]=max(dp[j][k]+1){0<=k<j&&a[k]==a[i]}但这样复杂度是O^3

但是dp[j][k]中的k肯定是选择最靠近j的那个,假如有k1<k2<……<kx<j,dp[j][kx]>=dp[j][k(1~x-1)]

这样的话，复杂度就变成O^2了

#include <cstdio>
#include <cstring>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <stack>
typedef long long ll;
#define X first
#define Y second
#define mp(a,b) make_pair(a,b)
#define pb push_back
#define sd(x) scanf("%d",&(x))
#define Pi acos(-1.0)
#define sf(x) scanf("%lf",&(x))
#define ss(x) scanf("%s",(x))
#define maxn 10000000
#include <ctime>
const int inf=0x3f3f3f3f;
const long long mod=1000000007;
using namespace std;
int dp[5005][5005];
int num[5005];
int main()
{
    #ifdef local
    freopen("in","r",stdin);
    //freopen("data.txt","w",stdout);
    int _time=clock();
    #endif
    int n;
    cin>>n;
    for(int i=1;i<=n;i++)
        sd(num[i]);
    int ans=0;
    int la;
    for(int i=1;i<=n;i++)
    {
        la=0;
        for(int j=0;j<i;j++)
        {
            dp[i][j]=dp[j][la]+1;
            if(num[i]==num[j])la=j;
            ans=max(dp[i][j],ans);
        }
    }
    cout<<ans<<endl;
    #ifdef local
    printf("time: %d\n",int(clock()-_time));
    #endif
}