hdu 5064 Find Sequence（单调性优化DP）

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hdu 5064 Find Sequence

一开始设想先排序然后两两相减并找出LIS来得到结果，显然没有考虑周全

根据题解，这题需要用到单调性DP。DP[i][j]表示序列以下标为i和j结尾，转移方程为

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);}#elsetypedef __int64 LL;inline void opt64(LL a){    printf("%I64d", a);}#endifconst 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;}

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