立志用最少的代碼做最高效的表達

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PAT甲級最優(yōu)題解——>傳送門

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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.
?? 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, customer

Input Specification:
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 Specification:
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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題意：K個人，N個窗口，每個窗口分為黃線內(nèi)和黃線外， 每個窗口黃線內(nèi)可以站M個人。
最初按窗口順序在黃線里排隊，如果黃線里排滿，則統(tǒng)一排到黃線外（只有一個隊列）。
當(dāng)某一窗口有一個人辦理完業(yè)務(wù)，黃線外的第一個人立刻補充到該窗口黃線內(nèi)的隊伍里。

老老實實寫模擬即可。

解題思路：由于最后需要按編號查詢， 因此定義結(jié)構(gòu)體，元素有編號、所需時間、完成時間等等。然后定義結(jié)構(gòu)體隊列， 將時間簡化為分鐘，一分鐘一次循環(huán)， 接下來對每個隊列進行操作即可。

細(xì)節(jié)：

如果K<M*N， 那么入隊列的人以K為準(zhǔn)

如果到了17點，顧客還沒有被服務(wù)(注意是被服務(wù))。 那么輸出sorry

（PS：抱著Debug的決心敲得代碼，反反復(fù)復(fù)調(diào)試確認(rèn)了很久才提交代碼，結(jié)果一遍過，太開心了！）

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#include<bits/stdc++.h> using namespace std;struct costomer{int id, n_t1, n_t2, f_t; //編號、所需時間、實際完成時間 int h, m; //完成的小時數(shù)，分鐘數(shù) }; bool cmp(costomer c1, costomer c2) {return c1.id < c2.id; //id號升序排列 }int main() {int N, M, K, Q; cin >> N >> M >> K >> Q;queue<costomer>q[N];int id = 1;//處理前M*N個顧客 int Min = min(M, K); while(1) {for(int j = 0; j < N; j++) {costomer cos;int t; cin >> t;cos.id = id++;cos.n_t1 = cos.n_t2 = t;q[j].push(cos);if(id > min(M*N, K)) goto loop; //滿足條件則退出循環(huán) }}loop : ;int now_cos_num = max(K-M*N, 0); //剩余未處理的乘客 vector<costomer>fin; //存放結(jié)果 int t = 0;while(1) { //每次循環(huán)都是一分鐘 t++;int temp = 0; for(int i = 0; i < N; i++) { //每一分鐘所有窗口同時-1 if(!q[i].empty()) { //如果隊列非空 q[i].front().n_t1--;if(q[i].front().n_t1 == 0) { //如果剩余處理時間為0 q[i].front().f_t = t; //壓入結(jié)果隊列 costomer cos = q[i].front();q[i].pop();fin.push_back(cos); //結(jié)果序列存放該顧客if(now_cos_num-- > 0) { //如果還有剩余顧客 int x; cin >> x; //輸入，壓入隊列 cos.id = id++; cos.n_t1 = cos.n_t2 = x; cos.f_t = 0;q[i].push(cos);}}} else { //如果空了，代表沒有后續(xù)的人，temp++。 temp++; }}if(temp == N) break; //如果temp=窗口數(shù)，則說明無人，返回。 }sort(fin.begin(), fin.end(), cmp); for(int i = 0; i < Q; i++) {int x; cin >> x; x-=1;if(fin[x].f_t-fin[x].n_t2 >= 540) cout << "Sorry\n";else printf("%02d:%02d\n", fin[x].f_t/60+8, fin[x].f_t%60);}return 0; }

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耗時：

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????????——錯過落日余暉，請記得還有漫天星辰。

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