PAT 1014 Waiting in Line (模拟)

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 (*N**M*+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.
• Customeri 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 customers 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. *Custome**r*5 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 *custome**r*5 at 08:10.

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

思路

这题有个坑，如果有人在17：00之前到了窗口前，就必须完成此人的业务。即，17：00前开始服务的，就必须服务完。

我自己写了一组样例，方便大家调试自己的程序

INPUT
2 1 4 4
600 539 400 200
1 2 3 4
OUPUT
18:00
16:59
23:39
Sorry

代码

#include <stdio.h>
#include <string>
#include <stdlib.h>
#include <iostream>
#include <vector>
#include <string.h>
#include <algorithm>
#include <cmath>
#include <map>
#include <limits.h> 
using namespace std;
int N, M, K, Q;
int t[1000 + 10]; //记录每个人咨询的时间 
vector<int> myv[30]; // 模拟每个窗口前的人的标号 1-K 
int pos = 1; //要插入到黄线内的人所在的位置 
int num = 0; //结束的人数 
int now = 0; // 当前的时间 
bool cmp1(vector<int> a, vector<int> b){
    return a.size() < b.size();
}

int main() {
    cin >> N >> M >> K >> Q;
    for(int i = 1; i <= K; i++){
        cin >> t[i];
    }
    while(num < K){
        // 进入黄线 
        vector<int> *min_t = min_element(myv + 1, myv + N + 1, cmp1);
        while(pos <= K && (*min_t).size() != M){
            int vi = min_t - myv;
            myv[vi].push_back(pos);
            //cout << vi << " " << pos << " " << now << endl;
            min_t = min_element(myv + 1, myv + N + 1, cmp1);
            pos++;
        }
        // 咨询业务 
        int flag = true;
        while(flag){
            if(num == K)    break;
            now++;
            for(int i = 1; i <= N; i++){
                if(myv[i].size()){
                    t[myv[i][0]]--;
                    if(t[myv[i][0]] == 0){
                        flag = false;   //有人完成，需要插入人 
                        t[myv[i][0]] = now;
                        num++;
                        myv[i].erase(myv[i].begin(), myv[i].begin() + 1);
                        //防止样例中，在540（17：00）刚刚结束时，再从第1变成第0 
                        if(now == 540){
                            for(int z = 0; z < myv[i].size(); z++){
                                t[myv[i][z]] = INT_MAX;
                            } 
                        }
                    }
                }
            }
            if(now >= 540)  break;
        }
        if(now >= 540){
            //把每个队伍的第一个咨询结束，不是第一个的无法完成咨询业务（INT_MAX） 
            for(int i = 1; i <= N; i++){
                for(int j = 0; j < myv[i].size(); j++){
                    if(j == 0 && t[myv[i][0]] != INT_MAX){
                        t[myv[i][0]] += now;
                    }
                    else{
                        t[myv[i][j]] = INT_MAX;
                    }
                }
            }
            // 还未插入到黄线内部的人，也无法完成咨询业务 
            while(pos <= K){
                t[pos++] = INT_MAX;
            }
            break;
        }
    }
    for(int i = 1; i <= Q; i++){
        int q = 0;
        cin >> q;
        if(t[q] == INT_MAX) cout << "Sorry" << endl;
        else printf("%02d:%02d\n", 8 + t[q] / 60, t[q] % 60);
    }
    return 0; 
}