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2014
01-16

九度-1326-Waiting in Line-模拟题

里面来自:2011年九度互动社区IT名企招聘上机考试热身赛

题目1326:Waiting in Line

题目描述:
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.

输入:
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.

输出:
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.

样例输入:
2 2 7 5
1 2 6 4 3 534 2
3 4 5 6 7
样例输出:
08:07
08:06
08:10
17:00
Sorry

题目大意:

有K个客户去银行办理业务,黄线内可以排队M个人。黄线外的人会优先选择人数少的队去排。 每个人花费的时间是不一样的。问每个人业务结束的时间。稍微有些繁琐的模拟。即使是17点之前开始办业务,17点之后仍然没有办完就是sorry。

 

#include <stdio.h>
#include <iostream>
#include <string>
#include <vector>
#include <map>
#include <queue>
#include <algorithm>

using namespace std;

bool _gDebug;
#define _P(a)  if (_gDebug) cout << #a":" << a << " "
#define _P2(a, b)   if (_gDebug) cout << #a":" << a << " "#b":" << b << endl
#define _P3(a, b, c) if (_gDebug) {\
    cout << #a":" << a << " "#b":" << b << " "#c":" << c << endl;}

#define INF 100000
queue<pair<int, int> > wd[30];
int ans[1005];
int N, M, K, Q;
int ki;
int tt;

void insert(int id, int t) {
    static int i = 0;
    _P3(i, id, t);
    wd[i++].push(make_pair(id, t));
    if (i == N) i = 0;
}
int getMin() {
    int mid = 0;
    int mt = wd[0].front().second;
    for (int i = 1; i < N; ++i) {
        if (wd[i].front().second < mt) {
            mt = wd[i].front().second;
            mid = i;
        }
    }
    for (int i = 0; i < N; ++i) {
        wd[i].front().second -= mt;
        _P3(i, wd[i].front().first, wd[i].front().second);
    }
    tt += mt;
    ans[wd[mid].front().first] = tt;
    wd[mid].pop();
    return mid;
}

int main(int argc, char *argv[]) {
    if (argc > 1) _gDebug = true;
    cin >> N >> M >> K >> Q;
    int tmp;
    for (ki = 0; ki<K && ki<N*M; ++ki) {
        cin >> tmp;
        insert(ki, tmp);
    }
    while (ki < K) {
        int mid = getMin();
        cin >> tmp;
        _P3(ki, mid, tmp);
        wd[mid].push(make_pair(ki, tmp));
        ++ki;
    }
    _P2(tmp, tt);
    for (int i = 0; i < N; ++i) {
        tmp =  tt;
        while (! wd[i].empty()) {
            tmp += wd[i].front().second;
            ans[wd[i].front().first] = tmp;
            wd[i].pop();
        }
    }
    for (int i = 0; i < Q; ++i) {
        cin >> tmp;
        tmp = ans[tmp-1];
        if (tmp <= 9*60) {
            int hh = 8 + tmp/60; 
            int mm = tmp % 60;
            printf("%02d:%02d\n", hh, mm);
        } else {
            printf("Sorry\n");
        }
    }

    return 0;
}

 


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  2. 我还有个问题想请教一下,就是感觉对于新手来说,递归理解起来有些困难,不知有没有什么好的方法或者什么好的建议?