HDU4122-Alice's mooncake shop

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Alice's mooncake shop

                                                                           Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
                                                                                                      Total Submission(s): 4124    Accepted Submission(s): 1071


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”(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<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 10Jan 1 2000 9 105 220 20 20 10 1087 9 5 100 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
 

Recommend
lcy


题意:有家月饼店,接到n个做月饼的订单,但月饼店只有在从2000年一月一日0点为第一个小时的前m个小时内做月饼,而且只能在整点的时候做月饼,并且做月饼不花费时间,也就是一瞬间就可以做超级多的月饼,而每个月饼有t个小时的保质期,每个月饼保存每小时需要花费s元,而在可以做月饼的m小时内,不同小时做月饼花费的钱也不同,每个订单都有交付订单的时间,某年某月某日某时交付该订单所要求的r个月饼,求完成所有订单所花费的最少的钱的数目。

解题思路:把时间统一成从起点时间开始的小时数,可以看成在一条时间轴上,n个订单就是n个特定结点,就是要找出每个结点前t个时间点内最小的花费。用单调队列类似维护范围内最小的单个月饼的花费,由于这个花费除了本身的价格还与距离订单时间的时间长短有关,因为保存月饼还要花费,而且与保存的时间越长花费越多,所以队列里要存储两个变量,价格和时间。


#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <algorithm>#include <cmath>#include <queue>#include <vector>#include <set>#include <stack>#include <map>#include <climits>using namespace std;#define LL long longconst int INF=0x3f3f3f3f;struct node{    LL x,t;}b;queue<node>q;LL qq[100090][2];char ch[10];int a[15]={0,31,28,31,30,31,30,31,31,30,31,30,31};int check(LL x){    if((x%4==0&&x%100!=0)||x%400==0) return 1;    else return 0;}LL check_month(){    if (!strcmp(ch,"Jan")) return 1;    else if (!strcmp(ch,"Feb")) return 2;    else if (!strcmp(ch,"Mar")) return 3;    else if (!strcmp(ch,"Apr")) return 4;    else if (!strcmp(ch,"May")) return 5;    else if (!strcmp(ch,"Jun")) return 6;    else if (!strcmp(ch,"Jul")) return 7;    else if (!strcmp(ch,"Aug")) return 8;    else if (!strcmp(ch,"Sep")) return 9;    else if (!strcmp(ch,"Oct")) return 10;    else if (!strcmp(ch,"Nov")) return 11;    else return 12;}LL check_t(){    LL h,y,d,ans=0,m;    scanf("%s %lld %lld %lld",ch,&d,&y,&h);    m=check_month();    for(LL i=2000;i<y;i++)    {        if(!check(i)) ans+=365*24;        else ans+=366*24;    }    for(LL i=1;i<m;i++)    {        if(i==2&&check(y)) ans+=29*24;        else ans+=a[i]*24;    }    ans+=(d-1)*24;    return ans+h;}int main(){    LL n,m,t,s,x;    while(~scanf("%lld %lld",&n,&m)&&(n+m))    {        int head=0,rear=0;        for(int i=1; i<=n; i++)        {            b.t=check_t();            scanf("%lld",&b.x);            q.push(b);        }        scanf("%lld %lld",&t,&s);        LL ans=0;        for(int i=0;i<m;i++)        {            scanf("%lld",&x);            while(head<rear&&x<=qq[rear-1][0]+(i-qq[rear-1][1])*s) rear--;            while (head<rear&&i>qq[head][1]+t) head++;            qq[rear][0]=x,qq[rear++][1]=i;            while(!q.empty()&&i==q.front().t)            {                ans+=((i-qq[head][1])*s+qq[head][0])*q.front().x;                q.pop();            }        }        printf("%lld\n",ans);    }    return 0;}

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