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

理清思路还是可以的，这道题一开始想用树状数组做，尝试了下发现2个测试点超时，由于没有注意到查询输入是按时间排序的，因此每次输出不必从头计数，可以从之前计数点继续，然后就是对无效数据的处理，这里用到了两个map记录同车牌的上一个状态，与状态为"in"的下标用于修改数据是否有效；

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <map>
#include <iterator>
#define MAX 10005
#define IN 1
#define OUT 2
using namespace std;

struct car{
    bool invalid;
    char p_num[10];
    char status[5];
    int h, m, s, time;
};

car c[MAX];
int N, K;
int maxtime = 0;
map<string, int> car_cost;

bool cmp(const car &a, const car &b)
{
    return a.time < b.time;
}

void init()
{
    scanf("%d %d", &N, &K);
    for(int i=0; i<N; i++)
    {
        scanf("%s %d:%d:%d %s", c[i].p_num, &c[i].h, &c[i].m, &c[i].s, c[i].status);
        c[i].time = c[i].h*3600+c[i].m*60+c[i].s;
    }
    sort(c, c+N, cmp);
}

void check()
{
    map<string, int> car_status;
    map<string, int> car_index;
    for(int i=0; i<N; i++)
    {
        string car = c[i].p_num;
        string sta = c[i].status;
        if(sta[0] == 'i')
        {
            c[i].invalid = true;//in只有碰到out才是有效数据，所以首先全部置为无效
            car_status[car] = IN;
            car_index[car] = i;//记录下标
        }
        else if(sta[0] == 'o')
        {
            if(car_status[car] == IN)//前一个为in时则这一对数据有效，修改前一个in有效
            {
                c[car_index[car]].invalid = false;
                car_cost[car] += c[i].time - c[car_index[car]].time;
                if(maxtime < car_cost[car])
                {
                    maxtime = car_cost[car];
                }
                car_status[car] = OUT;//修改状态，否则下一个out也会匹配
            }
            else{//该数据无效
                c[i].invalid = true;
            }
        }
    }
}

int main()
{
    //freopen("in.txt", "r", stdin);
    int h, m, s, time;
    int j = 0;
    init();
   /* for(int i=0; i<N; i++)
    {
        printf("%s %d:%d:%d %s\n", c[i].p_num, c[i].h, c[i].m, c[i].s, c[i].status);
    }*/
    check();
    int total = 0;
    for(int i=0; i<K; i++)
    {
        scanf("%d:%d:%d", &h, &m, &s);
        time = h*3600+m*60+s;
        for(; c[j].time<=time && j<N; j++)
        {
            if(!c[j].invalid)
            {
                if(c[j].status[0] == 'i')
                    total++;
                else
                    total--;
            }
        }
        printf("%d\n", total);
    }
    for(map<string, int>::iterator it=car_cost.begin(); it!=car_cost.end(); it++)//map遍历默认key从小到大，因此直接输出即可
    {
        if(it->second == maxtime)
            cout << it->first << " ";
    }
    h = maxtime/3600;
    m = (maxtime-3600*h)/60;
    s = maxtime-3600*h-60*m;
    printf("%02d:%02d:%02d", h, m, s);
    return 0;
}