Monotonic queue optimization of multiple knapsacks

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When solving multiple knapsacks, our simplest method is to treat all items as independent and solve the 01 knapsack. The time complexity is $O(NV\sum K)$ , then we can use binary optimization, the time complexity is$O(NVlog\sum k)$ , but we still can't get through some problems. At this time, we need to use monotonic queues to optimize.

We first consider the ordinary state transition equation: $f[i][j]=max(f[i-1][j-k\ast w]+k\ast v),(j-k\ast w>=0)$
Then we can consider adjusting the loop to optimize the one-dimensional array, that is, to ensure that each time is transformed from the previous time.
Obtain the equation$f[j]=max(pre[j-k\ast w]+k\ast v),(j-k\ast w>=0)$

Then we find that $f[j]\; can$ only be determined by$pre[j-k\ast w]$ transformed.
We put$f\left[m\right]$ (m is the total capacity of the backpack) grouping:

what we need to get is$max(pre[j-k\ast w]+k\ast v)$ , so we maintain a$max\left(pre\right[j-k\ast w]+k\ast$The maximum value of$v$$)$ .
Then we can get:

we find that for f[j+k*v], k is incremented by 1 each time, and w is required to be added to the value of each subsequent item, so we need to do some conversion.

Code:

#include<bits/stdc++.h>
#define endl '\n'
#define null NULL
#define ls p<<1
#define rs p<<1|1
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define ll long long
//#define int long long
#define pii pair<int,int>
#define ull unsigned long long
#define pdd pair<double,double>
#define lowbit(x) x&-x
#define all(x) (x).begin(),(x).end()
#define sz(x) (int)(x).size()
#define IOS ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
char *fs,*ft,buf[1<<20];
#define gc() (fs==ft&&(ft=(fs=buf)+fread(buf,1,1<<20,stdin),fs==ft))?0:*fs++;
inline int read()
{
    
    
    int x=0,f=1;
    char ch=gc();
    while(ch<'0'||ch>'9')
    {
    
    
        if(ch=='-')
            f=-1;
        ch=gc();
    }
    while(ch>='0'&&ch<='9')
    {
    
    
        x=x*10+ch-'0';
        ch=gc();
    }
    return x*f;
}
using namespace std;
const int N=2e4+55;
const int inf=0x3f3f3f3f;
const int mod=998244353;
const double eps=1e-6;
const double PI=acos(-1);
int q[N],f[N],pre[N];
void solve()
{
    
    
    int n,m;
    cin>>n>>m;
    for(int i=1;i<=n;i++)
    {
    
    
        int v,w,s;
        cin>>v>>w>>s;
        memcpy(pre,f,sizeof f);
        for(int j=0;j<v;j++)
        {
    
    
            int head=0,tail=-1;
            for(int k=j;k<=m;k+=v)
            {
    
    
                if(head<=tail&&(k-q[head])/v>s)
                    head++;
                if(head<=tail)
                    f[k]=max(f[k],pre[q[head]]+(k-q[head])/v*w);
                while(head<=tail&&pre[q[tail]]-(q[tail]-j)/v*w<=pre[k]-(k-j)/v*w)
                    tail--;
                q[++tail]=k;
            }
        }
    }
    cout<<f[m]<<endl;
}
signed main()
{
    
    
//    int t;
//    cin>>t;
//    while(t--)
    solve();
    return 0;
}