1095 Cars on Campus （30 分)

2019-04-15

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 (≤104), the number of records, and K (≤8×104) the number of queries. Then N lines follow, each gives a record in the format:

1	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 outrecord. 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

作者: CHEN, Yue

单位: 浙江大学

时间限制: 300 ms

内存限制: 64 MB

代码长度限制: 16 KB

题目大意

给出车进出浙大的列表，求给定时间校园内车的数量和在校园时间最长的车

分析

首先读取输入，根据inout配对筛选合法的数据。把进入时间和出校时间存放至两个vector。再读入查询的时间，根据upper_bound函数查找，作差即为校园内车数。时间复杂度nlgn。一辆车可以来回进出校园，所以统计停车时间时需要都加上。

代码

#include <iostream>
#include <vector>
#include <stack>
#include <algorithm>
#include <deque>
#include <map>
#include <set>
#include <cstring>
#include <cmath>
#include <sstream>

using namespace std;
int n,k;
struct sta{
    string plate;
    int time;
    string in;
    sta(const string &plate, int time, const string &in) : plate(plate), time(time), in(in) {}
    sta() {}
};

vector<sta> v;
vector<int> intime;
vector<int> outime;
map<string,int> carpark;
vector<string> longcar;
bool cmp(sta s1,sta s2){
    return s1.plate!=s2.plate?s1.plate<s2.plate:s1.time<s2.time;
}
int main() {
    cin>>n>>k;
    string p,in;
    int h,m,s;
    for(int i=0;i<n;i++){
        cin>>p;
        scanf("%d:%d:%d ",&h,&m,&s);
        cin>>in;
        int x=h*3600+m*60+s;
        sta st(p,x,in);
        v.push_back(st);
    }
    sort(v.begin(),v.end(),cmp);
    int maxt=-1;
    for(int i=0;i<v.size()-1;i++){
        if(v[i].plate==v[i+1].plate&&v[i].in=="in"&&v[i+1].in=="out"){
            carpark[v[i].plate]+=v[i+1].time-v[i].time;
            int x=carpark[v[i].plate];
            if(x>maxt)
            {
                maxt=x;
                longcar.clear();
                longcar.push_back(v[i].plate);
            }else if(x==maxt){
                longcar.push_back(v[i].plate);
            }
            intime.push_back(v[i].time);
            outime.push_back(v[i+1].time);
        }
    }
    sort(intime.begin(),intime.end());
    sort(outime.begin(),outime.end());
    sort(longcar.begin(),longcar.end());
    int cnt=0;
    for(int i=0;i<k;i++){
        scanf("%d:%d:%d",&h,&m,&s);
        int x=h*3600+m*60+s;
        int ii=upper_bound(intime.begin(),intime.end(),x)-intime.begin();
        int oi=upper_bound(outime.begin(),outime.end(),x)-outime.begin();
        cnt=ii-oi;
        printf("%d\n",cnt);
    }
    for(string s:longcar)
        cout<<s<<" ";
    printf("%02d:%02d:%02d",maxt/3600,(maxt%3600)/60,maxt%60);
    return 0;
}

其他

这题我做复杂了，根据车进出时间排序，再判断是in或者out可以做到o(n)的复杂度。不过nlng也能过。