HDU 3401 Trade dp+单调队列优化

题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=3401

Trade

Time Limit: 2000/1000 MS (Java/Others)
Memory Limit: 32768/32768 K (Java/Others)
#### 问题描述
> 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?

输入

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.

输出

The most money lxhgww can earn.

样例输入

1

5 2 0

2 1 1 1

2 1 1 1

3 2 1 1

4 3 1 1

5 4 1 1

样例输出

3

题意

一个人，一开始有无数的钱和0张股票，接下来的t天里，在第i天，他能够选择以api的价格买进一张股票，且最多允许买asi张；或者以bpi的价格卖出股票，且最多卖出bpi张。任何时刻手头的股票不能超过maxp张，且任意两次交易需要隔至少w天。问最多能赚多少钱。

题解

dp[i][j]表示第i天持有j张股票。

则易知:dp[i][j]=max{dp[i-1][j],dp[i-w-1][k]-(j-k)*api(买入),dp[i-w-1][k]+(k-j)*bpi)(卖出)}

因为当j固定时只需找出max{dp[i-w-1][k]+k*api(或者bpi)}，由于有asi(或者bsi)的限制，所以需要用单调队列来维护一下(否则只需要维护一个最大值就ok了)。所以复杂度是O^2。

代码

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<ctime>
#include<vector>
#include<cstdio>
#include<string>
#include<bitset>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
using namespace std;
#define X first
#define Y second
#define mkp make_pair
#define lson (o<<1)
#define rson ((o<<1)|1)
#define mid (l+(r-l)/2)
#define sz() size()
#define pb(v) push_back(v)
#define all(o) (o).begin(),(o).end()
#define clr(a,v) memset(a,v,sizeof(a))
#define bug(a) cout<<#a<<" = "<<a<<endl
#define rep(i,a,b) for(int i=a;i<(b);i++)
#define scf scanf
#define prf printf
typedef __int64 LL;
typedef vector<int> VI;
typedef pair<int,int> PII;
typedef vector<pair<int,int> > VPII;
const int INF=0x3f3f3f3f;
const LL INFL=0x3f3f3f3f3f3f3f3fLL;
const double eps=1e-8;
const double PI = acos(-1.0);
//start----------------------------------------------------------------------
const int maxn=2222;
int n,maxp,w;
int ap[maxn],bp[maxn],as[maxn],bs[maxn];
///dp[i][j]表示第i天持j张股票
int dp[maxn][maxn];
///单调队列
PII que[maxn];
int main() {
    int tc;
    scf("%d",&tc);
    while(tc--){
        scf("%d%d%d",&n,&maxp,&w);
        for(int i=0;i<n;i++){
            scf("%d%d%d%d",&ap[i],&bp[i],&as[i],&bs[i]);
        }
        ///初始化
        for(int i=0;i<n;i++){
            for(int j=0;j<=maxp;j++){
                dp[i][j]=-INF;
            }
        }
        for(int j=0;j<=min(maxp,as[0]);j++){
            dp[0][j]=-ap[0]*j;
        }
        for(int i=1;i<n;i++){
            ///第i天不进行交易
            for(int j=0;j<=maxp;j++){
                dp[i][j]=dp[i-1][j];
            }
            ///第i天进行交易
            if(i-w-1<0){
                for(int j=0;j<=min(maxp,as[i]);j++){
                    dp[i][j]=max(dp[i][j],-ap[i]*j);
                }
                continue;
            }
            ///买入
            int f=1,r=1;
            for(int j=0;j<=maxp;j++){
                int x=dp[i-w-1][j]+ap[i]*j;
                while(f<r&&que[r-1].X<x) r--;
                que[r++]=mkp(x,j);
                while(f<r&&que[f].Y+as[i]<j) f++;
                dp[i][j]=max(dp[i][j],que[f].X-j*ap[i]);
            }
            ///卖出
            f=r=1;
            for(int j=maxp;j>=0;j--){
                int x=dp[i-w-1][j]+bp[i]*j;
                while(f<r&&que[r-1].X<x) r--;
                que[r++]=mkp(x,j);
                while(f<r&&que[f].Y-bs[i]>j) f++;
                dp[i][j]=max(dp[i][j],que[f].X-j*bp[i]);
            }
        }
        int ans=0;
        for(int i=0;i<=maxp;i++){
            ans=max(ans,dp[n-1][i]);
        }
        prf("%d\n",ans);
    }
    return 0;
}
//end-----------------------------------------------------------------------