PAT A1095 Cars on Campus

题目描述：

　　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 (≤10?4??), the number of records, and K (≤8×10?4??) 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

参考代码：

 1 /****************************************************
 2 PAT A1095 Cars on Campus
 3 ****************************************************/
 4 #include <iostream>
 5 #include <algorithm>
 6 #include <vector>
 7 #include <map>
 8
 9 using namespace std;
10
11 struct ParkRecord {
12     string num;
13     int time;
14     string state;
15 };
16
17 bool cmpByNumAndState(ParkRecord a, ParkRecord b) {
18     return (a.num == b.num ? a.time < b.time : a.num < b.num);
19 }
20
21 bool ccmpByTime(ParkRecord a, ParkRecord b) {
22     return a.time < b.time;
23 }
24
25 int main() {
26     int recordCnt = 0, askTime = 0, maxParkTime = -1;
27
28     cin >> recordCnt >> askTime;
29
30     map<string, int> ParkTime;
31     vector<ParkRecord> EffectRecord;
32     vector<ParkRecord> OriginRecord(recordCnt);
33
34     //读入原始的记录
35     int hour = 0, min = 0, sec = 0, time = 0;
36     for (int i = 0; i < recordCnt; ++i) {
37         cin >> OriginRecord[i].num;
38         scanf("%d:%d:%d", &hour, &min, &sec);
39         cin >> OriginRecord[i].state;
40
41         time = hour * 3600 + min * 60 + sec;
42         OriginRecord[i].time = time;
43     }
44
45     //对原始记录进行排序
46     sort(OriginRecord.begin(), OriginRecord.end(), cmpByNumAndState);
47
48     //将有效的记录存入新的容器中
49     for (int i = 0; i < OriginRecord.size() - 1; ++i) {
50         if (OriginRecord[i].num == OriginRecord[i + 1].num &&
51             OriginRecord[i].state == "in" && OriginRecord[i + 1].state == "out") {
52             EffectRecord.push_back(OriginRecord[i]);
53             EffectRecord.push_back(OriginRecord[i + 1]);
54
55             if (ParkTime[OriginRecord[i].num] == 0) {
56                 ParkTime[OriginRecord[i].num] = 0;
57             }
58
59             ParkTime[OriginRecord[i].num] += OriginRecord[i + 1].time - OriginRecord[i].time;
60
61             if (maxParkTime < ParkTime[OriginRecord[i].num]) {
62                 maxParkTime = ParkTime[OriginRecord[i].num];
63             }
64         }
65     }
66
67     //对有效记录进行排序
68     sort(EffectRecord.begin(), EffectRecord.end(), ccmpByTime);
69
70     //输出查询时间点的车辆数目
71     int now = 0, carCnt = 0;
72     for (int i = 0; i < askTime; ++i) {
73         scanf("%d:%d:%d", &hour, &min, &sec);
74
75         time = hour * 3600 + min * 60 + sec;
76
77         for (; now < EffectRecord.size() && EffectRecord[now].time <= time; ++now) {
78             EffectRecord[now].state == "in" ? carCnt++: carCnt--;
79         }
80
81         cout << carCnt << endl;
82     }
83
84     //输出停车时间最长的车辆牌号和最长时间
85     for (map<string, int>::iterator itr = ParkTime.begin(); itr != ParkTime.end(); ++itr) {
86         if (itr->second == maxParkTime) {
87             cout << itr->first << ‘ ‘;
88         }
89     }
90
91     printf("%02d:%02d:%02d", maxParkTime / 3600, (maxParkTime % 3600) / 60, maxParkTime % 60);
92
93     return 0;
94 }

注意事项：

　　1：在Visual Studio 2019中不能使用scanf(会被认为不安去，无法通过编译)，但是在最终判定的时候会因为scanf_f产生编译错误，因此提交的时候记得改成scanf。

原文地址：https://www.cnblogs.com/mrdragon/p/11651459.html