题目地址https://www.patest.cn/contests/pat-a-practise/1014

1014. Waiting in Line (30)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

Suppose a bank has N windows open for service. There is a yellow line in front of the windows which devides the waiting area into two parts. The rules for the customers to wait in line are:

• The space inside the yellow line in front of each window is enough to contain a line with M customers. Hence when all the N lines are full, all the customers after (and including) the (NM+1)st one will have to wait in a line behind the yellow line.
• Each customer will choose the shortest line to wait in when crossing the yellow line. If there are two or more lines with the same length, the customer will always choose the window with the smallest number.
• Customer[i] will take T[i] minutes to have his/her transaction processed.
• The first N customers are assumed to be served at 8:00am.

Now given the processing time of each customer, you are supposed to tell the exact time at which a customer has his/her business done.

For example, suppose that a bank has 2 windows and each window may have 2 custmers waiting inside the yellow line. There are 5 customers waiting with transactions taking 1, 2, 6, 4 and 3 minutes, respectively. At 08:00 in the morning, customer₁ is served at window₁ while customer₂ is served at window₂. Customer₃ will wait in front of window₁ and customer₄ will wait in front of window₂. Customer₅ will wait behind the yellow line.

At 08:01, customer₁ is done and customer₅ enters the line in front of window₁ since that line seems shorter now. Customer₂ will leave at 08:02, customer₄ at 08:06, customer₃ at 08:07, and finally customer₅ at 08:10.

Input

Each input file contains one test case. Each case starts with a line containing 4 positive integers: N (<=20, number of windows), M (<=10, the maximum capacity of each line inside the yellow line), K (<=1000, number of customers), and Q (<=1000, number of customer queries).

The next line contains K positive integers, which are the processing time of the K customers.

The last line contains Q positive integers, which represent the customers who are asking about the time they can have their transactions done. The customers are numbered from 1 to K.

Output

For each of the Q customers, print in one line the time at which his/her transaction is finished, in the format HH:MM where HH is in [08, 17] and MM is in [00, 59]. Note that since the bank is closed everyday after 17:00, for those customers who cannot be served before 17:00, you must output "Sorry" instead.

Sample Input

2 2 7 51 2 6 4 3 534 23 4 5 6 7

Sample Output

08:0708:0608:1017:00Sorry

注：代码中的fprintf为调试信息，输出到标准错误流，提交时并影响，可以直接提交。

写之前没有多想，有些代码重复率比较高，可以打包成函数，代码风格懒得改了，仅供参考

#include<iostream>#include<string.h>using namespace std;/*每个窗口的服务队伍采用环状数组，pn【i】指向第i个窗口当前被服务的node，例如一个队伍目前有3个人，1、2、3，pn指向1；当第四个人到来的时候，队伍变成了423，pn指向2；当都五个人来的时候，队伍变成了453，pn指向3*/typedef struct cus{//顾客节点int dur, start, end, window;//每个顾客的服务时间、开始时间，截止时间，在那个窗口进行服务} cus;typedef struct node{//队伍节点int nage;//服务的是哪个顾客int endtime;//服务这个顾客的结束时间} node;cus cu[1000];node lines[10][20];int n, m, k, q;// k customers ,q querys, n windows, m customers/windowint pn[20];//point to every lineint qr[1000];int select();int main(){cin >> n >> m >> k >> q;for (int i = 0; i < k; i++){cin >> cu[i].dur;}for (int i = 0; i < q; i++){cin >> qr[i];}memset(pn, 0, sizeof(int)*20);memset(lines, 0, sizeof(node)* 200);for (int i = 0; i < m*n && i<k; i++){//如果顾客数量少于黄线内可以容纳的总量cu[i].window = i%n;lines[pn[i%n]][i%n].nage = i;// record the customer numberif (pn[i%n] == 0){lines[pn[i%n]][i%n].endtime = lines[m - 1][i%n].endtime + cu[i].dur;//这个顾客结束服务的时间=上个顾客结束的时间+这个顾客服务的时长fprintf(stderr,"t1 : %d, t2 : %d, t3 : %d\n", lines[pn[i%n]][i%n].endtime, lines[m - 1][i%n].endtime, cu[i].dur);cu[i].start = lines[m - 1][i%n].endtime;//这个顾客开始被服务的时间}else{lines[pn[i%n]][i%n].endtime = lines[pn[i%n] - 1][i%n].endtime + cu[i].dur;cu[i].start = lines[pn[i%n] - 1][i%n].endtime;fprintf(stderr, "t1 : %d, t2 : %d, t3 : %d\n", lines[pn[i%n]][i%n].endtime, lines[pn[i%n] - 1][i%n].endtime, cu[i].dur);}cu[i].end = cu[i].start + cu[i].dur;pn[i%n] = (pn[i%n] + 1) % m;//服务指针移动}/*下面的代码除了for循环的条件和window的选择外，和上面一模一样*/for (int i = m*n; i < k; i++){//最开始位于黄线外等待的顾客cu[i].window = select();//选择一个合适的窗口lines[pn[cu[i].window]][cu[i].window].nage = i;// record the customer numberif (pn[cu[i].window] == 0){lines[pn[cu[i].window]][cu[i].window].endtime = lines[m - 1][cu[i].window].endtime + cu[i].dur;cu[i].start = lines[m - 1][cu[i].window].endtime;fprintf(stderr, "t1 : %d, t2 : %d, t3 : %d\n", lines[pn[cu[i].window]][cu[i].window].endtime, lines[m - 1][cu[i].window].endtime, cu[i].dur);}else{lines[pn[cu[i].window]][cu[i].window].endtime = lines[pn[cu[i].window] - 1][cu[i].window].endtime + cu[i].dur;cu[i].start = lines[pn[cu[i].window] - 1][cu[i].window].endtime;fprintf(stderr, "t1 : %d, t2 : %d, t3 : %d\n", lines[pn[cu[i].window]][cu[i].window].endtime, lines[pn[cu[i].window] - 1][cu[i].window].endtime, cu[i].dur);}cu[i].end = cu[i].start + cu[i].dur;pn[cu[i].window] = (pn[cu[i].window] + 1) % m;}//下面为调试信息for (int i = 0; i < k; i++){fprintf(stderr, "%d %d %d %d\n", cu[i].start, cu[i].dur, cu[i].end, cu[i].window);}//end of debugfor (int i = 0; i < q; i++){if (cu[qr[i]-1].start >= 9 * 60){printf("Sorry\n");}else{int hour = cu[qr[i] - 1].end / 60 + 8;int minute = cu[qr[i] - 1].end % 60;printf("%02d:%02d\n", hour, minute);}}return 0;}int select(){//从当前所有的队伍中选择一个最早有空位的窗口int min = -1;int ans = 0;for (int i = 0; i < n; i++){if (lines[pn[i]][i].endtime < min || min == -1){ans = i;min = lines[pn[i]][i].endtime;}}return ans;}