[FROM LUOGU]P2569 [SCOI2010]股票交易

P2569 [SCOI2010]股票交易

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SOL
设$dp[i][j]$表示第$i$天持有$j$股的最大收益

$30pts$：朴素转移$O(n^4)$
不买卖：$dp[i][j]=dp[i-1][j]$
买：如果是第一次买：$dp[i][j]=-j\ast ap[i](0<=j<=as[i])$
否则：$dp[i][j]=dp[i-k][j-t]-t\ast ap[i](k>=max(i-w-1,0),0<=t<=as[i])$
卖：$dp[i][j]=dp[i-k][j+t]+t\ast bp[i](k>=max(i-w-1,0),0<=t<=bs[i])$
上述三者取$Max$

$50pts$：$O(n^3)$
考虑到有些天可以选择不买，于是$i$直接由$i-w-1$转移过来即可

$100pts$：$O(n^2)$
考虑到会取$max$，自然想到用单调队列优化
买：$dp[i][j]=max(dp[i-w-1][j-t]-t\ast ap[i])(0<=t<=as[i])$
令$k=j-t$
$dp[i][j]=max(dp[i-w-1][k]-(j-k)\ast ap[i])=max(dp[i-w-1][k]+k\ast ap[i])-j\ast ap[i]$
卖：$dp[i][j]=max(dp[i-w-1][j+t]+t\ast bp[i]0(0<=t<=bs[i])$
令$k=j+t$
$dp[i][j]=max(dp[i-w-1][k]+(k-j)\ast bp[i])=max(dp[i-w-1][k]+k\ast bp[i])-j\ast bp[i]$

代码：

#include<bits/stdc++.h>
using namespace std;
#define re register
inline char nc(){
	static char buf[10000],*p1=buf,*p2=buf;
	return p1==p2&&(p2=(p1=buf)+fread(buf,1,10000,stdin),p1==p2)?EOF:*p1++;
}
inline int rd(){
	int re data=0;static char ch=0;ch=nc();
	while(!isdigit(ch))ch=nc();
	while(isdigit(ch))data=(data<<1)+(data<<3)+(ch^48),ch=nc();
	return data;
}
template<typename T>inline void chmax(T &a,T b){return a>b?a:b;}
const int N=2005;
int ap[N],bp[N],dp[N][N],ans,as[N],bs[N],q[N],l,r,t,w,m;
signed main(){
	t=rd(),m=rd(),w=rd();
	for(int re i=1;i<=t;++i)ap[i]=rd(),bp[i]=rd(),as[i]=rd(),bs[i]=rd();
	memset(dp,0xb3,sizeof(dp)),dp[0][0]=0; 
	for(int re i=1;i<=t;++i){
		for(int re j=0;j<=as[i];++j)dp[i][j]=-j*ap[i];
		for(int re j=0;j<=m;++j)dp[i][j]=max(dp[i][j],dp[i-1][j]);
		if(i<=w+1)continue;
		l=1,r=0;
		for(int re j=0;j<=m;++j){
		int re k=i-w-1;
		while(l<=r&&q[l]<j-as[i])++l;
		while(l<=r&&dp[k][j]+j*ap[i]>dp[k][q[r]]+q[r]*ap[i])--r;
		q[++r]=j,dp[i][j]=max(dp[i][j],dp[k][q[l]]-ap[i]*(j-q[l]));
		}
		l=1,r=0;
		for(int re j=m;j>=0;--j){
		int re k=i-w-1;
		while(l<=r&&q[l]>j+bs[i])++l;
		while(l<=r&&dp[k][j]+j*bp[i]>dp[k][q[r]]+q[r]*bp[i])--r;
		q[++r]=j,dp[i][j]=max(dp[i][j],dp[k][q[l]]+bp[i]*(q[l]-j));
		}
	}
	for(int re j=0;j<=m;++j)ans=max(ans,dp[t][j]);
	printf("%d",ans);
	return 0;
}