HDU-3401-Trade-dp-单调队列优化

HDU-3401-Trade-dp-单调队列优化

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【Description】

Recently, lxhgww is addicted to stock, he finds some regular patterns after a few
days' study.
He forecasts the next T days' stock market. On the i'th day, you can buy one stock
with the price APi or sell one stock to get BPi.
There are some other limits, one can buy at most ASi stocks on the i'th day and at
most sell BSi stocks.
Two trading days should have a interval of more than W days. That is to say,suppose
you traded (any buy or sell stocks is regarded as a trade)on the i'th day, the next
trading day must be on the (i+W+1)th day or later.
What's more, one can own no more than MaxP stocks at any time.

Before the first day, lxhgww already has infinitely money but no stocks, of course
he wants to earn as much money as possible from the stock market. So the question
comes, how much at most can he earn? 

【Input】

The first line is an integer t, the case number.
The first line of each case are three integers T , MaxP , W .
(0 <= W < T <= 2000, 1 <= MaxP <= 2000) .
The next T lines each has four integers APi，BPi，ASi，BSi
( 1<=BPi<=APi<=1000,1<=ASi,BSi<=MaxP), which are mentioned above. 

【Output】

The most money lxhgww can earn. 

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【Examples 】

Sample Input

1
5 2 0
2 1 1 1
2 1 1 1
3 2 1 1
4 3 1 1
5 4 1 1

Sample Output

3

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【Problem Description】

 知道未来T天的股票买入价格和卖出价格，在满足以下条件下求所能获得的最大收益。
 1、一次买入或卖出操作后，必须间隔W天才能进行第二次操作。
 2、每天买入和卖出的股票数有限制。

【Solution】

 dp，单调队列优化
 定义dp数组dp[i][j],表示第i天买j只股票所能获得的最大收益。
 则买入的状态转移方程：dp[i][j]=max(dp[i][j],dp[i-W-1][k]-(j-k)*Ap[i])
                             =max(dp[i][j],dp[i-W-1][k]+k*Ap[i]-j*Ap[i])
   卖出的状态转移方程：dp[i][j]=max(dp[i][j],dp[i-W-1][k]+(k-j)*Bp[i])
                             =max(dp[i][j],dp[i-W-1][k]+k*Bp[i]-j*Bp[i])
 无论是买入或卖出，dp[i-W-1][k]+k*Xp[i]是跟k相关的值，j*Xp[i]是与j相关的值
 而且题目要求是最大收益，对于同一个j，枚举所有的k求dp[i-W-1][k]+k*Xp[i]的最大
 值就好了，那么就可以将这个式子通过单调队列来维护。
 注意最后求值的时候，还要保证abs(k-j)<=Xs[i]，即买入或卖出的数量不超过限制。

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【Code】

/*
 * @Author: Simon
 * @Date: 2018-08-17 09:12:27
 * @Last Modified by:   Simon
 * @Last Modified time: 2018-08-17 09:35:57
 */
#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define maxn 2015
int Api[maxn]; int Bpi[maxn];
int Asi[maxn]; int Bsi[maxn];
int f[maxn][maxn];
int head,tail;
int T, MaxP, W;
struct node
{
    int val,num;
}q[maxn*2];
void buy(int i,int j)
{
    while(head<tail&&q[tail-1].val<f[i-W-1][j]+j*Api[i]) tail--;//dp[i][k]+k*Api[i];
    q[tail].val=f[i-W-1][j]+j*Api[i];
    q[tail++].num=j;
    while(head==-1||(head<tail&&j-q[head].num>Asi[i])) head++;//保证买的数目不超过限制
    f[i][j]=max(f[i][j],q[head].val-j*Api[i]);//dp[i][j]=max(dp[i][j],dp[i][k]-(j-k)*Api[i])
}
void sale(int i,int j)
{
    while(head<tail&&q[tail-1].val<f[i-W-1][j]+j*Bpi[i]) tail--;//dp[i][k]+k*Bpi[i]
    q[tail].val=f[i-W-1][j]+j*Bpi[i];
    q[tail++].num=j;
    while(head==-1||(head<tail&&q[head].num-j>Bsi[i])) head++;//保证卖的数目不超过限制
    f[i][j]=max(f[i][j],q[head].val-j*Bpi[i]);//dp[i][j]=max(dp[i][j],dp[i][k]+(k-j)*Bpi[i])
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int t;
    cin >> t;
    while (t--)
    {
        cin >> T >> MaxP >> W;
        for (int i = 1; i <= T; i++) cin >> Api[i] >> Bpi[i] >> Asi[i] >> Bsi[i];
        for(int i=0;i<=T;i++)
            for(int j=0;j<=MaxP;j++) f[i][j]=-1e8;

        for(int i=1;i<=T;i++) f[i][0]=0;//任意一天不买不卖所获得的收益为0
        for(int i=1;i<=W+1;i++)
        {
            for(int j=0;j<=Asi[i];j++)
            {
                f[i][j]=-j*Api[i];//前w+1天，只能买进
            }
        }
        f[0][0]=0;
        for(int i=2;i<=T;i++)
        {
            head=0,tail=0;
            for(int j=0;j<=MaxP;j++) f[i][j]=max(f[i-1][j],f[i][j]);//不买不卖时的最大值
            if(i<=W+1) continue;
            for(int j=0;j<=MaxP;j++) buy(i,j);//因为是买进，所以比应该正序，保证比当前买的数目小的状态都已算过
            head=0,tail=0;
            for(int j=MaxP;j>=0;j--) sale(i,j);//相反...
        }
        int ans=-1e8;
        for(int j=0;j<=MaxP;j++) ans=max(ans,f[T][j]);
        cout<<ans<<endl;
    }
    cin.get(), cin.get();
    return 0;
}