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;
}