1017 Queueing at Bank (25)

1017 Queueing at Bank (25)
Suppose a bank has K windows open for service. There is a yellow line in front of the windows which devides the waiting area into two parts. All the customers have to wait in line behind the yellow line, until it is his/her turn to be served and there is a window available. It is assumed that no window can be occupied by a single customer for more than 1 hour.

Now given the arriving time T and the processing time P of each customer, you are supposed to tell the average waiting time of all the customers.

Input Specification:

Each input file contains one test case. For each case, the first line contains 2 numbers: N (<=10000) - the total number of customers, and K (<=100) - the number of windows. Then N lines follow, each contains 2 times: HH:MM:SS - the arriving time, and P - the processing time in minutes of a customer. Here HH is in the range [00, 23], MM and SS are both in [00, 59]. It is assumed that no two customers arrives at the same time.

Notice that the bank opens from 08:00 to 17:00. Anyone arrives early will have to wait in line till 08:00, and anyone comes too late (at or after 17:00:01) will not be served nor counted into the average.

Output Specification:

For each test case, print in one line the average waiting time of all the customers, in minutes and accurate up to 1 decimal place.

Sample Input:

7 3
07:55:00 16
17:00:01 2
07:59:59 15
08:01:00 60
08:00:00 30
08:00:02 2
08:03:00 10
Sample Output:

8.2
解题思路：
排队问题，计算等待时间。
1.将时间转化成秒，按照到达时间排序。
2.记录每个窗口的结束时间，从中选择最早结束的，让等待的客户前往。

#include<cstdio>
#include<vector>
#include<algorithm>
using namespace std;
const int maxn=111;
const int inf=1e8;
struct Custom{
    int cometime,p;
}cus;
vector<Custom>custom;
int change(int h,int m,int s){
    return h*3600+m*60+s;
}
bool cmp(Custom a,Custom b){
    return a.cometime <b.cometime ;
}
int endtime[maxn];  //记录每个窗口当前服务结束时间
int main() {
    int c,w,totaltime=0;
    int sttime=change(8,0,0);
    int edtime=change(17,0,0);
    scanf("%d%d",&c,&w);
    for(int i=0;i<w;i++)
        endtime[i]=sttime;
    for(int i=0;i<c;i++){
        int h,m,s,p;
        scanf("%d:%d:%d %d",&h,&m,&s,&p);
        int cometime=change(h,m,s);
        if(cometime>edtime) continue;
        cus.cometime =cometime;
        cus.p=p<=60?p*60:3600;
        custom.push_back(cus) ;
    }
    sort(custom.begin() ,custom.end() ,cmp);
    for(int i=0;i<custom.size() ;i++){
        int idx=-1,minendtime=inf;
        for(int j=0;j<w;j++){
            if(endtime[j]<minendtime){
                minendtime=endtime[j];
                idx=j;
            }
        }
        if(endtime[idx]<=custom[i].cometime)
            endtime[idx]=custom[i].cometime+custom[i].p;
        else{
            totaltime+=(endtime[idx]-custom[i].cometime);
            endtime[idx]+=custom[i].p;
        }
    }
    if(custom.size() ==0) printf("0.0");
    else printf("%.1f",totaltime/60.0/custom.size() );
    return 0;
}