hdu 3401

设f[i][j]为第i天有j张股票时的最大收益

易得f[i][j]=max{f[i-w+1]+k*ap[i]-j*ap[i](j<k<=j+bs[i]),f[i-w+1]+k*bp[i]-j*bp[i](j-as[i]<=k<j),f[i-1][j]}

这样用单调队列维护f[i-w+1]+k*a(b)p[i]的最大值就能降低转移复杂度

每次写DP代码总是很丑。。。

//#include<bits/stdc++.h>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<queue>
#define inc(i,l,r) for(int i=l;i<=r;i++)
#define dec(i,l,r) for(int i=l;i>=r;i--)
#define link(x) for(edge *j=h[x];j;j=j->next)
#define mem(a) memset(a,0,sizeof(a))
#define inf 1e9
#define ll long long
#define succ(x) (1<<x)
#define NM 2000+5
using namespace std;
int read(){
    int x=0,f=1;char ch=getchar();
    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
    while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
    return x*f;
}
int T,n,m,w,ans,f[NM][NM],q[NM],qh,qt,_x,_y,x,y;
int main(){
//    freopen("data.in","r",stdin);
    T=read();
    while(T--){
        n=read();m=read();w=read();ans=0;mem(f);mem(q);
        inc(i,1,m)f[0][i]=-inf;
        inc(i,1,w){
            x=read();y=read();_x=read();_y=read();
            inc(j,0,m)f[i][j]=f[i-1][j];
            inc(j,0,min(_x,m))
            f[i][j]=max(f[i][j],-j*x);
//            inc(j,1,m)printf("%d ",f[i][j]);printf("
");
        }
        inc(i,w+1,n){
            y=read();x=read();_y=read();_x=read();
            qh=1;qt=0;mem(q);
            inc(j,0,m)f[i][j]=f[i-1][j];
            dec(j,m,1){
                while(qh<=qt&&f[i-w-1][q[qt]]+x*q[qt]<=f[i-w-1][j]+x*j)qt--;
                while(qh<=qt&&q[qh]>j+_x-1)qh++;
                q[++qt]=j;int t=q[qh];
                f[i][j-1]=max(f[i][j-1],f[i-w-1][t]+x*t-(j-1)*x);
            }//sell
            qh=1;qt==0;mem(q);
            inc(j,0,m-1){
                while(qh<=qt&&f[i-w-1][q[qt]]+y*q[qt]<=f[i-w-1][j]+y*j)qt--;
                while(qh<=qt&&q[qh]<j-_y+1)qh++;
                q[++qt]=j;int t=q[qh];
                f[i][j+1]=max(f[i][j+1],f[i-w-1][t]+y*t-(j+1)*y);
            }//buy
//            inc(j,0,m)printf("%d ",f[i][j]);printf("
");
        }
        inc(i,0,m)ans=max(ans,f[n][i]);
        printf("%d
",ans);
    }
    return 0;
}