1095 Cars on Campus (30分)

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:ssrepresents the time point in a day by hour:minute:second, with the earliest time being 00:00:00and 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

题意：查询校园内还有多少车辆，输出本天内在校园内停放最久的车的牌号，如果不止一辆，按字母递增顺序输出。

思路：最难的是数据清洗，最终数据目标：记录为偶数条，按车牌和时间排序后，每辆车的两条记录前一条是in、后一条是out。对于清洗数据我们需要先对数据按车牌和时间排序，用迭代器（切记）遍历，如果本记录和下一条记录为一辆车&&本记录是in&&下条记录是out，则跌代价直接跳过这两条记录，反之，删除本条记录。这样就已经清洗完毕了。接下来需要用map来求停留时间最长的车，相当简单。查询每个时刻校园内还有多少辆车，因为时刻是递增的，所以这个思路可以参照1093题，通过计数的方式来获取。

#include<vector>

#include<iostream>
#include<algorithm>
#include<unordered_map>
using namespace std;
struct node
{
    string plate;
    int time;
    bool is_out;
};
bool cmp(node a, node b){
    return a.plate != b.plate ? a.plate < b.plate : a.time < b.time;
}
bool cmp_time(node a, node b){
    return a.time < b.time;
}
int main(){
    int n, k;
    cin >> n >> k;
    vector<node> vec(n);
    unordered_map<string, int> un_map;
    for(int i = 0; i < n; i++){
        string s, status;
        node temp;
        int hour, min, sen;
        cin >> s;
        scanf("%d:%d:%d", &hour, &min, &sen);
        cin >> status;
        temp.plate = s;
        temp.time = hour * 3600 + min * 60 + sen;
        if(status == "in"){
            temp.is_out = false;
        }else
        {
            temp.is_out = true;
        }
        vec[i] = temp;
        
        

    }
    sort(vec.begin(), vec.end(), cmp);
    
    vector<int> query(k);
    for(int i = 0; i < k; i++){
        int hour, min, sen;
        scanf("%d:%d:%d", &hour, &min, &sen);
        int total_time = hour * 3600 + min * 60 + sen;
        query[i] = total_time;
    }

    // for(int i = 0; i < vec.size(); i++){

    //     cout << vec[i].plate << "====" << vec[i].time << "====" << vec[i].is_out << endl;
    // }
    vector<node>::iterator it = vec.begin();
    while (it != vec.end())
    {
        if(it -> plate == (it + 1) -> plate && it -> is_out == 0 && (it + 1) -> is_out == 1){
            it = it + 2;
        }else
        {
            vec.erase(it);
            
        }
        

    }
    int max_time = -1;
    string out_s = "";
    for(int i = 0; i < vec.size() / 2; i++){
        un_map[vec[i * 2].plate] += (vec[i * 2 + 1].time - vec[i * 2].time);
        if(max_time < un_map[vec[i * 2].plate]){

            max_time = un_map[vec[i * 2].plate];

            out_s = vec[i * 2].plate;


        }else if(max_time == un_map[vec[i * 2].plate])
        {
            out_s += " ";
            out_s += vec[i * 2].plate;
        }
        
        
    
    }

    

    sort(vec.begin(), vec.end(), cmp_time);
    int flag = 0;
    int cnt_in = 0;
    int cnt_out = 0;
    for(int i = 0; i < vec.size(); i++){
        if(flag >= k){
            break;
        }
        if(vec[i].time <= query[flag]){
            if(vec[i].is_out == true){
                cnt_out++;
            }else
            {
                cnt_in++;
            }
            
        }else
        {
            printf("%d\n", cnt_in - cnt_out);
            flag++;
            i--;
        }

        


    }
    if(vec[vec.size() - 1].time <= query[k - 1]){
            printf("%d\n", cnt_in - cnt_out);
    }
    cout << out_s;
    printf(" %02d:%02d:%02d\n", max_time / 3600, (max_time % 3600) / 60, max_time % 60);


    return 0;
}