題目連結

題解

• 首先考慮直接$dp$的寫法$dp_{i,j,k}$表示在前$i$個數選擇了$j$個數後末尾連續未選擇的數個數為$k$可取得的最大值，轉移也很簡單。
• 但這個做法是$O(n^3)$的，觀察一下可以發現$k$這一維可以用單調佇列優化掉，然後複雜度就為$O(n^2)$
檢視程式碼

#include <bits/stdc++.h>
using namespace std;
#define _for(i,a,b) for(int i = (a);i <= (b);++i)
typedef long long ll;
const int maxn = 5000+5;
const int mod = 1e9+7;
ll qpow(ll a,ll b){ll res = 1;for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
int que[maxn][maxn];
int top[maxn],tail[maxn];
int a[maxn];
ll dp[maxn][maxn];

int main()
{
#ifndef ONLINE_JUDGE
    freopen("simple.in","r",stdin);
    freopen("simple.out","w",stdout);
#endif
    int n,m,k;
    cin>>n>>m>>k;
    for(int i = 1;i <= n;++i)scanf("%d",&a[i]);
    que[k][0] = 0;
    tail[k] = 1;
    memset(dp,-0x3f,sizeof(dp));
    dp[0][k]=0;
    ll ans = -1;
    for(int i = 1;i <= n;++i){
        for(int j = 0;j < k;++j){
            int &s = top[j+1];
            int &t = tail[j+1];
            while(s!=t&&i-que[j+1][s]>=m+1){
                s++;
            }
            if(s!=t)
            dp[i][j] = dp[que[j+1][s]][j+1]+a[i];
            int &s2 = top[j],&t2 = tail[j]; 
            while(s2!=t2&&dp[i][j]>=dp[que[j][t2-1]][j]){
                t2--;
            }
            que[j][t2++] = i;
            if(j==0&&n-i<m){
                ans = max(ans,dp[i][j]);
            }
            
        }
    }
    cout<<ans<<endl;
    return 0;
}