Jennie

一个比较水的动态规划

$dp[i][j]=max(dp[i][j],dp[i-1][k])+a_i*jquad kin [j-1,j-1+s]$

然后这个玩意可以用有限队列

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#define int long long
using namespace std;
int n,w,s;
int a[100001];
int dp[5005][5005];
deque<int> q;
int ans=-1008600110086001;
signed main(){
    scanf("%lld%lld%lld",&n,&w,&s);
    for(int i=1;i<=n;++i){
        scanf("%lld",&a[i]);
    }
    memset(dp,-0x3f,sizeof(dp));
    dp[0][0]=0;
    for(int i=1;i<=n;++i){
        int r=0;
        for(int j=1;j<=min(i,w);++j){
            while(!q.empty()&&q.front()<j-1){
                    q.pop_front();
                }
            while(r<=w&&r<=j-1+s){
                while(!q.empty()&&dp[i-1][q.back()]<=dp[i-1][r]){
                    q.pop_back();
                }
                q.push_back(r);
                r++;
            }
            dp[i][j]=dp[i-1][q.front()]+a[i]*j;
        }
        while(!q.empty())
        q.pop_back();
    }
    for(int i=1;i<=w;++i){
        ans=max(ans,dp[n][i]);
        //cout<<dp[n][i]<<endl;
    }
    cout<<ans;
    return 0;
}