hdu 5064 Find Sequence

DP[i][j] = max(DP[k][i] + 1) { k<=i && num[j] - num[i] >= num[i] - num[k] }

但是直接求解时间复杂度为O(n^3)。该状态方程具有特殊性质，即

若 k 满足 { k<=i && num[j] - num[i] >= num[i] - num[k] } ， 则 k 在 j+1 的情况下也满足。所以当计算 j+1 时，k 可直接从上一状态下继续进行。

#include <stdio.h>
#include <vector>
#include <string>
#include <algorithm>
#include <queue>
#include <cstring>
#include <map>
#include <set>
#include <iostream>
#include <cmath>
using namespace std;
#ifdef __GNUC__
typedef long long LL;
inline void opt64(LL a)
{
    printf("%lld", a);
}
#else
typedef __int64 LL;
inline void opt64(LL a)
{
    printf("%I64d", a);
}
#endif
const int MAXN = (1<<22)+5;
const int MAXM = 2050;
const double eps = 1e-10;
const double PI = acos(-1.0);
inline int cmp(const double x, const double y)
{
    return ((x-y)>eps) - ((x-y)<-eps);
}
inline int cmp(const double x)
{
    return ((x)>eps) - ((x)<-eps);
}
int n, m, mp[MAXN], mq[MAXM], cnt;
int dp[MAXM][MAXM];
int solve()
{
    int res = -1;
    if (n < 3) return n;
    sort(mp, mp+n);
    cnt = 0;
    mp[n] = -1;
    for (int i = 0, tp = 0; i< n; ++i)
    {
        tp++;
        if (mp[i] != mp[i+1])
        {
            mq[cnt] = mp[i], dp[cnt][cnt] = tp, res = max(res,tp), tp = 0, ++cnt;
        }
    }
    for (int i = 0; i< cnt; ++i)
    {
        int k = i, ans = dp[i][i];
        for (int j = i+1; j< cnt; ++j)
        {
            for (; k>=0 && mq[j]-mq[i]>=mq[i]-mq[k]; --k)
            {
                ans = max(ans, dp[k][i]);
            }
            dp[i][j] = ans+1;
            res = max(res, ans+1);
        }
    }
    return res;
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
#endif
    int t;
    scanf("%d", &t);
    while (t--)
    {
        scanf("%d%d", &n, &m);
        for (int i = 0; i< n; ++i)
            scanf("%d", mp+i);
        printf("%d\n", solve());
    }
    return 0;
}