HDU 4884 TIANKENG’s rice shop(模拟)——BestCoder Round #2

TIANKENG’s rice shop

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Problem Description

TIANKENG managers a pan fried rice shop. There are n kinds of fried rice numbered 1-n. TIANKENG will spend t time for once frying. Because the pan is so small, TIANKENG can fry k bowls of fried rice with same kind at most. Assuming that there are m customers coming to the shop, and we know the arriving time of each customer and the brand and number of the fried rice they need. Could you tell TIANKENG the departure time of every customer respectively? Pay attention that TIANKNEG will serve the customer who comes earlier and he will fry the rice as much as possible. Meanwhile, customers are in queue depending on their arriving time(the earlier they arrive, the more front they stand).

 

Input

The first line contains a positive integer T(T<=100), referring to T test cases.
For each test case, the first line has 4 positive integer n(1<=n<=1000), t(1<=t<=10), k(1<=k<=5), m(1<=m<=1000), then following m lines , each line has a time(the time format is hh:mm, 0<=hh<=23, 0<=mm<=59) and two positive integer id(1<=id<=n), num(1<=num<=10), which means the brand number of the fried rice and the number of the fried rice the customer needs.
Pay attention that two or more customers will not come to the shop at the same time, the arriving time of the customer will be ordered by the time(from early time to late time)

 

Output

For each test case print m lines, each line contains a time referring to the departure time of the customer. There is a blank line between two test cases.

 

Sample Input

32 1 4 208:00 1 509:00 2 12 5 4 308:00 1 408:01 2 208:02 2 22 5 4 208:00 1 108:04 1 1

 

Sample Output

08:0209:0108:0508:1008:1008:0508:10

 

Source

BestCoder Round #2

 

/****************************************************/

出题人的解题思路：

模拟题。当顾客来的时候，天坑如果不在炒饭，那么就给这个顾客炒饭。如果没有顾客到来，那么天坑会一直等待顾客到来，空闲的时候不会炒饭。如果在做的时候来客人需要相同的炒饭那么不会帮这人炒。

题意：有m个顾客来店里买炒饭，给你每个顾客到达店里的时间、想要买的炒饭的种类以及数量，问每个顾客尽早离开的时间。

有以下条件需要满足：①顾客先到先服务；②若开始制作种类i的炒饭时，等待的顾客中多人选种类i的可以同时得到满足；③一次制作炒饭最多k碗，每次炒会炒尽可能多的份数，但不会有多余的留着备用；④顾客信息是按时间先后顺序给出的。还有一点要注意的是：每次制作炒饭的那段时间是不接单的。

此题一看就是模拟题，思路简单，但是要考虑的细节较多

举个例子，比如每次可以炒5份，每次5分钟。(即k=5,t=5)

第一个顾客08:00进来，点了2份A，

第二个顾客08:04进来，点了3份A。

在08:00开始炒的话，由于这个时候第二个顾客还没进来，所以就只炒2份，第一个顾客在08:05离开，这时才炒第二个的3份，所以第二个离开时间是08:10。

同样是每次可以炒5份，每次5分钟。

第一个顾客08:00进来，点了6份A，

第二个顾客08:01进来，点了5份B，

第三个顾客08:02进来，点了4份A。

同样地，先炒5份给第一个，还差一份，这是已经是08:05了，第三个顾客也进来了，所以这时直接炒5份A（因为会尽可能多地炒），08：10第一个和第三个可以同时离开。接着才炒第二个的。

另外需要注意的一点是顾客到达店里的时间必定是同一天的，即不会出现比如前一个顾客到达时间为23:59，后一个顾客到达时间为00:04，但是顾客离开店的时间未必是同一天，所以最终还需要对时间做点处理

放上两组数据以供参考

Input

100
3 5 4 6
23:50 1 4
23:51 2 1
23:52 3 1
23:53 2 1
23:54 3 3
23:56 2 2

Output
23:55
00:00
00:05
00:00
00:05
00:10

Input
100
3 5 4 3
23:50 1 5
23:52 2 7
23:58 2 1

Output
00:00
00:10
00:10

#pragma comment(linker, "/STACK:1024000000,1024000000")#include<stdio.h>#include<string.h>#include<stdlib.h>#include<queue>#include<math.h>#include<vector>#include<map>#include<set>#include<stdlib.h>#include<cmath>#include<string>#include<algorithm>#include<iostream>#define exp 1e-10using namespace std;const int N = 1005;const int inf = 1000000000;const int mod = 1000000007;struct person{    int t,id,num,ans;}s[N];int main(){    int T,n,t,k,m,i,j,hh,mm,f=0,c,K,tem;    scanf("%d",&T);    while(T--)    {        if(f++)            puts("");        scanf("%d%d%d%d",&n,&t,&k,&m);        for(c=i=0;i<m;i++)        {            scanf("%d:%d%d%d",&hh,&mm,&s[i].id,&s[i].num);            s[i].t=hh*60+mm;            s[i].ans=-1;        }        for(c=s[0].t,i=0;i<m;i++)        {            if(s[i].ans!=-1)                continue;            K=(s[i].num+k-1)/k;            tem=max(c,s[i].t)+(K-1)*t;            //printf("**%02d:%02d**\n",tem/60%24,tem%60);            K*=k;            for(j=i;j<m&&K&&s[j].t<=tem;j++)            {                if(s[j].id==s[i].id)                {                    n=min(s[j].num,K);                    s[j].num-=n;                    K-=n;                }                if(s[j].num==0&&s[j].ans==-1)                    s[j].ans=tem+t;            }            c=tem+t;        }        for(i=0;i<m;i++)            printf("%02d:%02d\n",s[i].ans/60%24,s[i].ans%60);    }    return 0;}

菜鸟成长记