标签：imu nal ensure empty min ide cat names code
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:
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, customer?1?? is served at window?1?? while customer?2?? is served at window?2??. Customer?3?? will wait in front of window?1?? and customer?4?? will wait in front of window?2??. Customer?5?? will wait behind the yellow line.
At 08:01, customer?1?? is done and customer?5?? enters the line in front of window?1?? since that line seems shorter now. Customer?2?? will leave at 08:02, customer?4?? at 08:06, customer?3?? at 08:07, and finally customer?5?? at 08:10.
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.
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.
2 2 7 5
1 2 6 4 3 534 2
3 4 5 6 7
08:07
08:06
08:10
17:00
Sorry
#include <stdio.h> #include <stdlib.h> #include <math.h> #include <algorithm> #include <iostream> #include <string.h> #include <queue> using namespace std; const int maxn = 1010; int n, m, k, q, now = 0; int t[maxn] = { 0 }; int done[maxn] = { 0 }; int pre[maxn] = { 0 }; struct people { int id; int time; int time_const; }; queue<people*> que[21]; int main() { scanf("%d %d %d %d", &n, &m, &k, &q); if (n == 0 || m == 0) { for (int i = 0; i < q; i++) { printf("Sorry\n"); } } for (int i = 1; i <= k; i++) { int time; scanf("%d", &time); people *p = new people; p->id = i; p->time = time; p->time_const = time; if(i<=n*m)que[(i-1)%n].push(p); else { int first = 99999999, first_j = 0; for (int j = 0; j < n; j++) { if (que[j].front()->time < first) { first = que[j].front()->time; first_j = j; } } for (int j = 0; j < n; j++) { que[j].front()->time -= first; } people* tmp_p = que[first_j].front(); que[first_j].pop(); now += first; done[tmp_p->id] = now; que[first_j].push(p); pre[tmp_p->id] = now - tmp_p->time_const; delete(tmp_p); } } for (int i = 0; i < n; i++) { int tmp_now = now; while (!que[i].empty()) { people* tmp_p = que[i].front(); que[i].pop(); tmp_now += tmp_p->time; pre[tmp_p->id] = tmp_now - tmp_p->time_const; done[tmp_p->id] = tmp_now; delete(tmp_p); } } for (int i = 0; i < q; i++) { int j; scanf("%d", &j); //printf("%d %d\n", pre[j], done[j]); if (pre[j]>=540)printf("Sorry\n"); else { printf("%02d:%02d\n", 8+done[j]/60,done[j]%60); } } system("pause"); }
注意点：考察队列知识，相对比较简单，有一个坑就是结束时间在17点后但开始时间17点前还是要输出时间的，一开始没注意就错了三个测试点。
PAT A1014 Waiting in Line （30 分）
标签：imu nal ensure empty min ide cat names code
原文地址：https://www.cnblogs.com/tccbj/p/10384598.html
