1014. Waiting in Line (30)

時間限制 400 ms 內存限制 65536 kB 代碼長度限制 16000 B 判題程序 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 5 1 2 6 4 3 534 2 3 4 5 6 7 Sample Output 08:07 08:06 08:10 17:00 Sorry

解析：該問題有兩種解決的方法，第一種就是完全模擬法，對所有的用戶進行便利模擬，計算出用戶離開的時間，這里需要注意的是17：00表示用戶接受服務的時間，不是離開的時間，也就是說只要在17：00或者之前接受服務都可以，就算該用戶的服務時間要一直持續到晚上十點。

Code:
/*************************************************************************> File Name: 1014.cpp> Author: > Mail: > Created Time: 2015年12月15日 星期二 21時05分54秒************************************************************************/#include<iostream> #include<cstring> #include<queue> #include<cstdio> using namespace std;#define INF 0x6fffffff// the waiting queue queue<int> wait[25]; queue<int> qu[25]; // the processing time int times[1010]; // the length of windows int timebase[25] = {0}; int leavetime[1010];int main(){int n,m,k,q;cin>>n>>m>>k>>q;int tmptime;for(int i=0; i<k; i++){cin>>tmptime;times[i] = tmptime;}// simulate the processint top = 0;int index;for(int i = 0; i<2*k; i++){int minlen = m;// if there is any customer not int the lineif(top != k){for(int j=0; j<n; j++){if(minlen > qu[j].size()){minlen = qu[j].size();index = j;}}}//find the queue have minimum lengthif(minlen != m){qu[index].push(top);wait[index].push(times[top]);top++;}// every queue is full or all customer in lineselse{long minwait = INF;bool empty = true;for(int j=0; j<n; j++){if(wait[j].empty()) continue;if(minwait > timebase[j]+wait[j].front()){minwait = timebase[j]+wait[j].front();index = j;empty = false;}}if(empty) break;timebase[index] += wait[index].front();leavetime[qu[index].front()] = timebase[index];qu[index].pop();wait[index].pop();}}int tmp;while(q--){cin>>tmp;tmp--;if(leavetime[tmp]-times[tmp] < 60*9){int hour = leavetime[tmp]/60;int min = leavetime[tmp]%60;printf("%02d:%02d\n",8+hour,min);}else{printf("Sorry\n");}}return 0; }

還有一種方法后續再寫。

?

轉載于:https://www.cnblogs.com/RookieCoder/p/5051144.html

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