POJ 4002 Alice's mooncake shop(DP,单调队列)

http://poj.org/problem?id=4002

Alice's mooncake shop
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 520   Accepted: 159

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. 

POJ 4002 Alice's mooncake shop(DP,单调队列)_第1张图片 

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

(0
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

“Jan 1 2000 9 10” means in Jan 1st 2000 at 9 o'clock , there's a consumer ordering 10 mooncakes. 
Maybe you should use 64-bit signed integers. The answer will fit into a 64-bit signed integer.

Source

Fuzhou 2011


题意:给出客户买月饼的时间和数量,每个时间点做月饼的成本,月饼的保质期和存储成本,求出最小成本。

求出每个时间点的最小成本即可,成本为min{制作成本+存储成本×存储时间},由于时间在变化,成本也在变化,但他们的“相对成本”之间的大小关系并没有发生变化,我们可以先预处理出“相对”于最后一天的成本(第i小时的成本为cost[i]+(M-i)*S),然后维护一个单调不减的队列,保证队首元素为最优即可。


#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#define sqr(x) ((x)*(x))
#define LL long long
#define itn int
#define INF 0x3f3f3f3f
#define PI 3.1415926535897932384626
#define eps 1e-10
#define mm 100007

using namespace std;

int q[mm],cost[mm],dp[mm];
int _f,_r;
int t[2501],r[2501];
int T,S,N,M;
int year,date,hour;
char _month[12][5]={ "Jan", "Feb", "Mar", "Apr", "May", "Jun", "Jul", "Aug", "Sep", "Oct", "Nov", "Dec" };
int dates[12]={31,29,31,30,31,30,31,31,30,31,30,31};
char month[5];

inline bool leap(int _y)
{
    return  (_y%400==0 || _y%100!=0 && _y%4==0);
}

inline int cal_hours(int _y,itn _m,int _d,int _h)
{
    int res=0;
    for (int i=2000;i<_y;i++)
    {
        if (leap(i))    res+=366;   else    res+=365;
    }

    if (leap(_y))   dates[1]=29;    else    dates[1]=28;

    for (int i=0;i<_m;i++)  res+=dates[i];

    res+=_d;
    res--;

    res*=24;

    return res+_h;
}

int main()
{
    #ifndef ONLINE_JUDGE
        freopen("/home/fcbruce/文档/code/t","r",stdin);
    #endif // ONLINE_JUDGE

    while (scanf("%d %d",&N,&M),N||M)
    {
        for (int i=0;icost[i])   _r--;
            q[++_r]=i;

            while (q[_f]


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