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hdu4122 Alice&#39;s mooncake shop 單調隊列

編輯：C++入門知識

hdu4122 Alice's mooncake shop 單調隊列

Alice's mooncake shop

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2908 Accepted Submission(s): 744

Problem 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.

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”(圓) 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
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 2011 Asia Fuzhou Regional Contest

題意：一個月餅店只在整點工作。在整點可以制作任意多個月餅。但是在不同的時間制作月餅花費的代價不一樣。而且如果一個月餅做好後儲存。每小時將花費S。但是儲存時間不超過T。現在給你該店工作的最長時間m。和n個訂單。問你完成訂單的最小花費。

思路：很簡單的單調隊列，把日期處理成小時後，用單調隊列維護下最小值就好，隊列頭加入新的時間點的，隊列尾刪除過期的。注意會有相同時間點的訂單。

/**
 * @author neko01
 */
//#pragma comment(linker, /STACK:102400000,102400000)
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include
using namespace std;
typedef long long LL;
#define min3(a,b,c) min(a,min(b,c))
#define max3(a,b,c) max(a,max(b,c))
#define pb push_back
#define mp(a,b) make_pair(a,b)
#define clr(a) memset(a,0,sizeof a)
#define clr1(a) memset(a,-1,sizeof a)
#define dbg(a) printf(%d
,a)
typedef pair pp;
const double eps=1e-8;
const double pi=acos(-1.0);
const int INF=0x7fffffff;
const LL inf=(((LL)1)<<61)+5;
char s[12][5]={Jan,Feb,Mar,Apr,May,Jun,Jul,Aug,Sep,Oct,Nov,Dec};
int day[12]={31,28,31,30,31,30,31,31,30,31,30,31};
struct node{
    int h,r;
}a[2555];
bool check(int y)
{
    return ((y%4==0&&y%100!=0)||y%400==0);
}
int input()
{
    char str[5];
    int d,y,h;
    scanf(%s%d%d%d,str,&d,&y,&h);
    int t=0,tmp=0,m=1;
    for(int i=2000;i