PAT(A) 1095. Cars on Campus (30)

Zhejiang University has 6 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 (<= 10000), the number of records, and K (<= 80000) 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

#include <cstdio>
#include <cstring>
#include <string>           //map<string, int> parkTime;
#include <map>
#include <algorithm>
using namespace std;
const int maxn=10010;

struct Car{
    char id[8];             //车牌号（注：比较大小时用 strcmp(,)
    int time;               //记录的 in or out 的时刻，以秒为单位
    char status[4];         //in or out
}all[maxn], valid[maxn];    //all为所有记录，valid为有效记录

//timeToInt(): 将时间转换为以秒为单位
int timeToInt(int hh, int mm, int ss){
    return hh*3600+mm*60+ss;
}
//cmp_byIdAndTime(): 先按车牌号字典序从小到大排序，相同的按时间从小到大排序
bool cmp_byIdAndTime(Car a, Car b){
    if( strcmp(a.id, b.id) )    return strcmp(a.id, b.id)<0;
    else    return a.time<b.time;
}
//cmp_byTime(): 按时间从小到大排序
bool cmp_byTime(Car a, Car b){
    return a.time<b.time;
}

int main()
{
    int n, k;           //n条记录，k个查询
    int hh, mm, ss;     //时，分，秒
    scanf("%d%d", &n, &k);
    for(int i=0; i<n; i++){
        scanf("%s %d:%d:%d %s", all[i].id, &hh, &mm, &ss, all[i].status);
        all[i].time=timeToInt(hh, mm, ss);  //转换为以秒为单位
    }
    //步骤(1): 按车牌号和时间排序
    sort(all, all+n, cmp_byIdAndTime);

    //步骤(2): 找有效记录，并记录某车牌号的最大总停留时间
    int num=0;                  //有效记录的条数（初始为0）
    int maxTime=-1;             //最长总停留时间（初始为-1）
    map<string, int> parkTime;  
    //车牌号 <-> 总停留时间（映射）：parkTime[string]=int  //#include <string>

    for(int i=0; i<n-1; i++){                   //遍历所有记录
        if( !strcmp(all[i].id, all[i+1].id) &&  //i和i+1是同一辆车
            !strcmp(all[i].status, "in")    &&  //i的记录为 in
            !strcmp(all[i+1].status, "out") )   //i+1的记录为 out
        {
            valid[num++]=all[i];                //i和i+1是配对的，将两个记录都存入valid[]
            valid[num++]=all[i+1];

            //划重点: map<string, int> parkTime;
            int inTime = all[i+1].time-all[i].time;         //记录此次停留时间
            if( parkTime.count(all[i].id) == 0 ){           //map中还没有这个车牌号
                parkTime[all[i].id] = 0;                    //置停留时间为零
            }
            parkTime[all[i].id] += inTime;                  //增加该车牌号的总停留时间
            maxTime = max(maxTime, parkTime[all[i].id]);    //更新最大总停留时间
        }
    }
    //步骤(3): 把有效记录按时间从小到大排序
    sort(valid, valid+num, cmp_byTime);

    //步骤(4): 输出该查询时刻校园内的车辆数
    //now 指向不超过当前查询时间的记录, numCar 为当前校园内车辆数
    int now=0, numCar=0;
    for(int i=0; i<k; i++){
        scanf("%d:%d:%d", &hh, &mm, &ss);
        int time=timeToInt(hh, mm, ss);
        //让now处理至当前当前查询时间
        while(now<num && valid[now].time<=time){
            if( !strcmp(valid[now].status, "in") )  numCar++;   //车辆进入
            else    numCar--;       //车辆离开
            now++;                  //now 指向下一条记录
        }
        printf("%d
", numCar);     //输出该时刻校园内的车辆数
    }

    //步骤(5): //输出所有最长总停留时间的车牌号
    //遍历所有车牌号
    map<string, int>::iterator it;
    for(it = parkTime.begin(); it != parkTime.end(); it++){
        if(it->second == maxTime){  　　//输出所有最长总停留时间的车牌号
            printf("%s ", it->first.c_str());
        }
    }
    //输出最长总停留时间
    printf("%02d:%02d:%02d
", maxTime/3600, maxTime%3600/60, maxTime%60);
    return 0;
}