1014 Waiting in Line (30 分)

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.
• Customeri​ will take Ti​ 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​. 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 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

鸣谢用户 不到76kg不改名 和 徐向荣 补充数据！

题目大意：有n个窗口，每个窗口最多有m个人，一共k个人，然后l个检查量，然后就是处理出来每个查询人的业务完成时间，每个人都会找到人数最少的去排，并且一旦排进去，就不能进入其他队伍里面。

思路：首先我们需要保留一个队伍中第一个人处理完成的时间，这样才知道后来的人要先进入哪个地方，同时我们还需要保留最后一个人的完成时间，这样就可以对新来的人算出来他对应的离开时间。

刚开始的时候a[index].pop()的位置写错了，应该是目前的最前面的那个人时间是否超过17.00，而不是第二个人的时间，这样会导致第一个人的时间无效。

也就是测试样例：

2 2 7 5
540 2 6 4 3 534 2
3 4 5 6 7

输出

Sorry
08:06
08:09
17:03
Sorry

 

#include <bits/stdc++.h>
using namespace std;
struct person{
    int poptime;//离开时间
    int bh;//编号
};
int main(){
    int n,m,k,l;
    scanf("%d%d%d%d",&n,&m,&k,&l);
    int i;
    int timeleft[n];//最后一个人的完成时间
    queue<person>a[n];
    int s=0;
    unordered_map<int,int>mymap;//结果统计
    memset(timeleft,0,sizeof(timeleft));
    bool sorry[k+1];//不能完成的人
    memset(sorry,0,sizeof(sorry));
    int j;
    for(i=0;i<k;i++){
        int temp;
        scanf("%d",&temp);
        if(i<n*m){
            if(a[s].empty()){
                a[s].push({temp,i+1});
                timeleft[s]=temp;
                s++;
                if(s==n)s=0;
            }
            else {
                a[s].push({timeleft[s]+temp,i+1});
                timeleft[s]+=temp;
                s++;
                if(s==n)s=0;
            }
        }
        else {
            int min=INT_MAX;
            int index=0;
            for(j=0;j<n;j++){//由于目前每个队伍都是满的，所以就查询第一个人谁先出去，就进入哪一个。
                if(a[j].front().poptime<min){min=a[j].front().poptime;index=j;}
            }
            mymap.insert(pair<int,int>({a[index].front().bh,a[index].front().poptime}));
            if(a[index].front().poptime>=540){//如果前面那个人办理业务时间超540，就代表后面的人都是sorry。
                a[index].pop();
                sorry[a[index].front().bh]=1;
            }
            else a[index].pop();
            a[index].push({timeleft[index]+temp,i+1});
            timeleft[index]+=temp;
        }
    }
    for(i=0;i<n;i++){//对剩余队列的人进行处理
        int time=0;
        while(!a[i].empty()){
            if(time>=540)sorry[a[i].front().bh]=1;
            mymap.insert(pair<int,int>(a[i].front().bh,a[i].front().poptime));
            time=a[i].front().poptime;
            a[i].pop();
        }
    }
    int num;
    for(i=0;i<l;i++){
        scanf("%d",&num);
        int hour=8+mymap[num]/60;
        int min=mymap[num]%60;
        if(sorry[num])printf("Sorry\n");
        else printf("%02d:%02d\n",hour,min);
    }
    system("pause");
}