P2517 [HAOI2010] Order (dp)

P2517 [HAOI2010] Order

Set $ ​​f [i] [j] $ denotes the $ i $ months, inventory is the minimum cost of $ j $

Enumeration month stock $ k $, then $ f [i] [j] = f [i-1] [k] + (j + U [i] -k) * D [i] + j * m , k <= min (j + U [i], S) $

Complexity $ O (nS ^ 2) $

The above equation removed items

$f[i][j]=(j+U[i])*D[i]+j*m+{f[i-1][k]-k*D[i]},k<=min(j+U[i],S)$

This can be directly followed by $ k $ $ j $ maintenance, do not open even monotonous queue

Complexity $ O (nS) $

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int n,m,S,f[53][10003],U[53],D[53],h[10003];
int main(){
    scanf("%d%d%d",&n,&m,&S);
    for(int i=1;i<=n;++i) scanf("%d",&U[i]);
    for(int i=1;i<=n;++i) scanf("%d",&D[i]);
    memset(f,63,sizeof(f)); f[0][0]=0;
    for(int i=1;i<=n;++i){
        int v=2e9,k=0;
        for(int j=0;j<=S;++j){
            while(k<=min(j+U[i],S)) v=min(v,f[i-1][k]-k*D[i]),++k;
            f[i][j]=v+(j+U[i])*D[i]+j*m;
        }
    }printf("%d",f[n][0]);
}