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

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提交代碼

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教訓：

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.

這句話的含義是如果不能在17:00之前開始服務，則輸出“Sorry”

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注意：vector就是可變長的動態數組，比較靈活好用

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加入元素：push_back()

刪除元素：用vector<int>::iterator ? it ?遍歷至要刪除的元素，然后v.erase(it)

元素個數：size()

訪問元素：和一般的數組一樣，直接訪問

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1 #include <cstdio> 2 #include <cstring> 3 #include <string> 4 #include <queue> 5 #include <vector> 6 #include <iostream> 7 using namespace std; 8 struct custom{ 9 int cost,finish; 10 }; 11 vector<int> v[21]; 12 custom cu[1005]; 13 int main(){ 14 //freopen("D:\\INPUT.txt","r",stdin); 15 int n,m,k,q; 16 scanf("%d %d %d %d",&n,&m,&k,&q); 17 int i,j; 18 for(i=0;i<k;i++){ 19 scanf("%d",&cu[i].cost); 20 } 21 for(i=0;i<n&&i<k;i++){ 22 cu[i].finish=cu[i].cost; 23 v[i].push_back(i); 24 } 25 for(;i<m*n&&i<k;i++){ 26 cu[i].finish=cu[v[i%n][v[i%n].size()-1]].finish+cu[i].cost; 27 v[i%n].push_back(i); 28 } 29 for(;i<k;i++){ 30 int mintime=cu[v[0][0]].finish,minnum=0; 31 for(j=1;j<n;j++){ 32 if(cu[v[j][0]].finish<mintime){ 33 minnum=j; 34 mintime=cu[v[j][0]].finish; 35 } 36 } 37 cu[i].finish=cu[v[minnum][v[minnum].size()-1]].finish+cu[i].cost; 38 vector<int>::iterator it=v[minnum].begin(); 39 v[minnum].erase(it); 40 v[minnum].push_back(i); 41 } 42 int num; 43 for(i=0;i<q;i++){ 44 scanf("%d",&num); 45 if(cu[num-1].finish-cu[num-1].cost<540){ 46 //cout<<num-1<<" "<<cu[num-1].finish<<endl; 47 int h=cu[num-1].finish/60+8; 48 if(h>17){ 49 continue; 50 } 51 int m=cu[num-1].finish%60; 52 if(h>9){ 53 cout<<h; 54 } 55 else{ 56 cout<<0<<h; 57 } 58 cout<<":"; 59 if(m>9){ 60 cout<<m; 61 } 62 else{ 63 cout<<0<<m; 64 } 65 cout<<endl; 66 } 67 else{ 68 cout<<"Sorry"<<endl; 69 } 70 } 71 return 0; 72 }

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轉載于:https://www.cnblogs.com/Deribs4/p/4708101.html

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