HDU 4122 Alice's mooncake shop 单调队列优化dp

时间：2022-07-10 11:19:50

Alice's mooncake shop

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://acm.hdu.edu.cn/showproblem.php?pid=4122

Description

The Mid-Autumn Festival, also known as the Moon Festival or Zhongqiu Festival is a popular harvest festival celebrated by Chinese people, dating back over 3,000 years to moon worship in China's Shang Dynasty. The Zhongqiu Festival is held on the 15th day of the eighth month in the Chinese calendar, which is in September or early October in the Gregorian calendar. It is a date that parallels the autumnal equinox of the solar calendar, when the moon is at its fullest and roundest.
HDU 4122 Alice's mooncake shop 单调队列优化dp
The
traditional food of this festival is the mooncake. Chinese family
members and friends will gather to admire the bright mid-autumn harvest
moon, and eat mooncakes under the moon together. In Chinese, “round”(圆)
also means something like “faultless” or “reuion”, so the roundest moon,
and the round mooncakes make the Zhongqiu Festival a day of family
reunion.

Alice has opened up a 24-hour mooncake shop. She always
gets a lot of orders. Only when the time is K o’clock sharp( K = 0,1,2
…. 23) she can make mooncakes, and We assume that making cakes takes no
time. Due to the fluctuation of the price of the ingredients, the cost
of a mooncake varies from hour to hour. She can make mooncakes when the
order comes,or she can make mooncakes earlier than needed and store them
in a fridge. The cost to store a mooncake for an hour is S and the
storage life of a mooncake is T hours. She now asks you for help to work
out a plan to minimize the cost to fulfill the orders.

 

Input

The input contains no more than 10 test cases.
For each test case:

The first line includes two integers N and M. N is the total number of orders. M is the number of hours the shop opens.

The next N lines describe all the orders. Each line is in the following format:

month date year H R

It means that on a certain
date, a customer orders R mooncakes at H o’clock. “month” is in the
format of abbreviation, so it could be "Jan", "Feb", "Mar", "Apr",
"May", "Jun", "Jul", "Aug", "Sep", "Oct", "Nov" or "Dec". H and R are
all integers.

All the orders are sorted by the time in increasing order.

The next line contains T and S meaning that the storage life
of a mooncake is T hours and the cost to store a mooncake for an hour is
S.

Finally, M lines follow. Among those M lines, the i th line( i starts from 1) contains a integer indicating the cost to make a mooncake during the i th hour . The cost is no more than 10000. Jan 1st 2000 0 o'clock belongs to the 1 st hour, Jan 1st 2000 1 o'clock belongs to the 2 nd hour, …… and so on.

(0<N <= 2500; 0 < M,T <=100000; 0<=S <= 200; R<=10000 ; 0<=H<24)

The input ends with N = 0 and M = 0.

 

Output

You should output one line for each test case: the minimum cost.

Sample Input

1 10 Jan 1 2000 9 10 5 2 20 20 20 10 10 8 7 9 5 10 0 0

Sample Output

70

HINT

题意

Alice开了家月饼店，现有2500笔订单，订单包括某小时（2000年1月1日0点算第1个小时的开始）和需要的月饼数量。然后给你前100000小时的信息，包括第i个小时做1个饼的花费cost[i]。然后给你月饼的保质期T（说明订单i只能买[order[i].hour-T ,order[i].hour ]这个区间生产的饼）和保存1小时的花费S，让你求最小的花费满足所有订单。

题解：

首先要把订单的时间转化成自2000年1月1日0点开始的第几小时，由于最多100000小时，所以最大到2012年的样子。然后维护一个最小值的单调队列。

具体实现：

首先我们已经获得第一个Order的单调队列，记为LQ，然后是处理第2个订单，我们把LQ分成2个部分。A：下标在order[2].hour-T, order[1].hour范围（A集合可能为空） B：下标在order[1].hour , order[2].hour的范围。对于A，由于第2个订单也可能使用到A集合里面的元素，所以我们要先对A集合所有元素加上1.2订单时间差的花费S*(order[2].hour- order[1].hour)（用于保存），然后再把B集合的元素加到单调队列里面，最后队列头的值即为第2个订单得到一个饼的最小花费，乘以订单数即可。之后的每个订单同样处理即可。注意，答案要long long。

http://altynai.me/2011/11/hdu-4121-4123/

代码

#include<iostream>

#include<stdio.h>

#include<vector>

#include<deque>

#include<queue>

#include<cstring>

using namespace std;

int n,m;

long long ans = ;

string mon[]={"", "Jan", "Feb", "Mar", "Apr", "May", "Jun", "Jul", "Aug", "Sep", "Oct", "Nov","Dec"};

int day[]={, , , , , , , , , , , , };

string M;

int d,y,h,num;

long long time[];

deque<pair<int,int> > QQ;

queue<pair<int,int> > Q;

int main()

{

    while(scanf("%d%d",&n,&m)!=EOF)

    {

        int tot = ;

        if(n==&&m==)break;

        while(!QQ.empty())QQ.pop_back();

        while(!Q.empty())Q.pop();

        memset(time,,sizeof(time));

        for(int i=;i<n;i++)

        {

            cin>>M;

            scanf("%d%d%d%d",&d,&y,&h,&num);

            for(int j=;j<y;j++)

            {

                if(( j %  ==  && j % ) || j %  ==  )d += ;

                else d += ;

            }

            int k;

            for(k=;k<=;k++)

                if(mon[k]==M)

                    break;

            for(int j=;j<k;j++)

            {

                if(j== &&( ( y %  ==  && y % ) || y %  ==  ))

                    d += ;

                else

                    d += day[j];

            }

            d--;

            Q.push(make_pair(num,d*+h));

        }

        int T,S;

        scanf("%d%d",&T,&S);

        long long ans = ;

        for(int i=;i<m;i++)

        {

            int x;

            scanf("%d",&x);

            while(!QQ.empty()&&x<=QQ.back().first + (i-QQ.back().second)*S)

                QQ.pop_back();

            QQ.push_back(make_pair(x,i));

            while(!Q.empty()&&i==Q.front().second)

            {

                while(!QQ.empty() && QQ.front().second + T < i )

                    QQ.pop_front();

                ans += (QQ.front().first + (i - QQ.front().second) * S) * Q.front().first;

                Q.pop();

            }

        }

        printf("%lld\n",ans);

    }

}

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