PAT_A1095#Cars on Campus

Source:

PAT A1095 Cars on Campus (30 分)

Description:

Zhejiang University has 8 campuses and a lot of gates. From each gate we can collect the in/out times and the plate numbers of the cars crossing the gate. Now with all the information available, you are supposed to tell, at any specific time point, the number of cars parking on campus, and at the end of the day find the cars that have parked for the longest time period.

Input Specification:

Each input file contains one test case. Each case starts with two positive integers N (≤), the number of records, and K (≤) the number of queries. Then N lines follow, each gives a record in the format:

plate_number hh:mm:ss status

where plate_number is a string of 7 English capital letters or 1-digit numbers; hh:mm:ss represents the time point in a day by hour:minute:second, with the earliest time being 00:00:00 and the latest 23:59:59; and status is either in or out.

Note that all times will be within a single day. Each in record is paired with the chronologically next record for the same car provided it is an out record. Any in records that are not paired with an out record are ignored, as are out records not paired with an in record. It is guaranteed that at least one car is well paired in the input, and no car is both in and out at the same moment. Times are recorded using a 24-hour clock.

Then K lines of queries follow, each gives a time point in the format hh:mm:ss. Note: the queries are given in ascending order of the times.

Output Specification:

For each query, output in a line the total number of cars parking on campus. The last line of output is supposed to give the plate number of the car that has parked for the longest time period, and the corresponding time length. If such a car is not unique, then output all of their plate numbers in a line in alphabetical order, separated by a space.

Sample Input:

16 7
JH007BD 18:00:01 in
ZD00001 11:30:08 out
DB8888A 13:00:00 out
ZA3Q625 23:59:50 out
ZA133CH 10:23:00 in
ZD00001 04:09:59 in
JH007BD 05:09:59 in
ZA3Q625 11:42:01 out
JH007BD 05:10:33 in
ZA3Q625 06:30:50 in
JH007BD 12:23:42 out
ZA3Q625 23:55:00 in
JH007BD 12:24:23 out
ZA133CH 17:11:22 out
JH007BD 18:07:01 out
DB8888A 06:30:50 in
05:10:00
06:30:50
11:00:00
12:23:42
14:00:00
18:00:00
23:59:00

Sample Output:

1
4
5
2
1
0
1
JH007BD ZD00001 07:20:09

Keys:

  • 模拟题

Code:

  1 /*
  2 Data: 2019-08-24 16:39:21
  3 Problem: PAT_A1095#Cars on Campus
  4 AC: 43:46
  5 
  6 题目大意：
  7 查询某时刻校园内停放车辆的数量，以及一天之中停留时间最长的车辆
  8 输入：
  9 第一行给出，记录数N<=1e4，查询数K<=8e4
 10 接下来N行，车牌号，时刻，出/入
 11 接下来K行，递增顺序给出查询时刻
 12 输出：
 13 各时刻校内停放车辆的数量；
 14 一天之中停留时间最长的车辆，id递增
 15 
 16 基本思路：
 17 设置两个列表；
 18 首先按车牌信息归类并按时间顺序递增排序，筛选有效信息，并统计各个车辆的停留时间
 19 再按照时间顺序重排有效信息，遍历列表并统计各时刻校内停放车辆的数量
 20 */
 21 #include<cstdio>
 22 #include<string>
 23 #include<vector>
 24 #include<iostream>
 25 #include<algorithm>
 26 using namespace std;
 27 struct node
 28 {
 29     string id;
 30     int time,status;
 31 }tp;
 32 vector<node> info,valid;
 33 
 34 bool cmp(const node &a, const node &b)
 35 {
 36     if(a.id != b.id)
 37         return a.id<b.id;
 38     else if(a.time != b.time)
 39         return a.time < b.time;
 40     else
 41         return a.status < b.status;
 42 }
 43 
 44 bool cmpTime(const node &a, const node &b)
 45 {
 46     return a.time < b.time;
 47 }
 48 
 49 int main()
 50 {
 51 #ifdef ONLINE_JUDGE
 52 #else
 53     freopen("Test.txt", "r", stdin);
 54 #endif // ONLINE_JUDGE
 55 
 56     int n,k;
 57     scanf("%d%d", &n,&k);
 58     for(int i=0; i<n; i++)
 59     {
 60         string s;
 61         int hh,mm,ss;
 62         cin >> s;
 63         tp.id =s;
 64         scanf("%d:%d:%d", &hh,&mm,&ss);
 65         tp.time=hh*3600+mm*60+ss;
 66         cin >> s;
 67         if(s=="in")
 68             tp.status = 0;
 69         else
 70             tp.status = 1;
 71         info.push_back(tp);
 72     }
 73 
 74     sort(info.begin(),info.end(),cmp);
 75     int maxTime=0;
 76     vector<string> longest;
 77     for(int i=0; i<info.size(); i++)
 78     {
 79         int sum=0;
 80         while(i+1<info.size() && info[i].id==info[i+1].id)
 81         {
 82             if(info[i].status==0 && info[i+1].status==1)
 83             {
 84                 sum += (info[i+1].time-info[i].time);
 85                 valid.push_back(info[i]);
 86                 valid.push_back(info[i+1]);
 87             }
 88             i++;
 89         }
 90         if(sum > maxTime)
 91         {
 92             maxTime = sum;
 93             longest.clear();
 94             longest.push_back(info[i].id);
 95         }
 96         else if(sum == maxTime)
 97             longest.push_back(info[i].id);
 98     }
 99 
100     sort(valid.begin(),valid.end(),cmpTime);
101     int keep=0,pt=0;
102     for(int i=0; i<k; i++)
103     {
104         int hh,mm,ss;
105         scanf("%d:%d:%d", &hh,&mm,&ss);
106         int time = hh*3600+mm*60+ss;
107         while(pt<valid.size() && valid[pt].time<=time)
108         {
109             if(valid[pt].status==0)
110                 keep++;
111             else
112                 keep--;
113             pt++;
114         }
115         printf("%d\n", keep);
116     }
117 
118     for(int i=0; i<longest.size(); i++)
119         printf("%s ", longest[i].c_str());
120     printf("%02d:%02d:%02d\n", maxTime/3600,(maxTime%3600)/60,maxTime%60);
121 
122     return 0;
123 }

 

posted @ 2019-07-12 21:34  林東雨  阅读(254)  评论(0)  编辑  收藏  举报