A1095. Cars on Campus

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<iostream>
#include<algorithm>
#include<string.h>
#include<string>
#include<map>
using namespace std;
typedef struct{
    char id[8];
    int time;
    int in;
}info;
bool cmp(info a, info b){
    if(strcmp(a.id, b.id) != 0)
        return strcmp(a.id, b.id) < 0;
    else return a.time < b.time;
}
bool cmp2(info a, info b){
    return a.time < b.time;
}
map <string, int> stay;
info data[10000], valid[10000];
int main(){
    int N, K;
    int hh, mm, ss, tt, maxTime = -1;
    char in_out[4]; 
    string id ;
    scanf("%d%d", &N, &K);
    for(int i = 0; i < N; i++){
        scanf("%s %d:%d:%d %s", data[i].id, &hh, &mm, &ss, in_out);
        data[i].time = hh * 3600 + mm * 60 + ss;
        data[i].in = strcmp(in_out, "in") == 0 ? 1 : 0;
    }
    sort(data, data + N, cmp);
    int cnt = 0;
    for(int i = 0; i < N - 1; i++){
        if(strcmp(data[i].id, data[i + 1].id) == 0 && data[i].in == 1 && data[i + 1].in == 0){
            id = data[i].id;
            if(stay.count(id) == 0)
                stay[id] = 0;
            stay[id] += (data[i + 1].time - data[i].time);
            if(stay[id] > maxTime)
                maxTime = stay[id];
            valid[cnt] = data[i];
            strcpy(valid[cnt].id, data[i].id);
            cnt++;
            valid[cnt] = data[i + 1];
            strcpy(valid[cnt].id, data[i + 1].id);
            cnt++;
        }
    }
    sort(valid, valid + cnt, cmp2);
    int pt = 0, cars = 0;
    for(int i = 0; i < K; i++){
        scanf("%d:%d:%d", &hh, &mm, &ss);
        tt = hh * 3600 + mm * 60 + ss;
        for( ; valid[pt].time <= tt && pt < cnt; pt++){
            if(valid[pt].in == 1)
                cars++;
            else cars--;
        }
        printf("%d\n", cars);
    }
    map<string, int> :: iterator it;
    for(it = stay.begin(); it != stay.end(); it++){
        if(it->second == maxTime)
            printf("%s ", it->first.c_str());
    }
    hh = maxTime / 3600; mm = maxTime % 3600 / 60; ss = maxTime % 60;
    printf("%02d:%02d:%02d",hh, mm, ss);
    cin >> N;
    return 0;
}

 

总结：

1、本题给出一堆车辆的出入信息，要求在任意查询时刻计算出校园内车辆数量，统计最终一天内停留时间最长的车辆。

      查询校园内车辆：与A1016电话费计算一样，给出的无序记录中有些记录无法配对，是无效记录，需要处理。可以用两个数组，data数组将所有数据读入，按照车牌id和时间综合排序后，进行筛选，只有两条紧邻的且id一致且前者为in后者为out的数据才是有效数据，将其存入valid数组。然后再对vaild数组仅仅按照时间排序。处理查询时，注意到所给的查询是按照时间顺序给出的，所以可以在前一次查询的基础上累计车辆个数以减小复杂度，依次遍历vaild直至达到所要查询的时间，在过程中out一辆车则减一，in一辆则加一。

　   统计停留时间最长的车：利用map存储车牌号与停留时间，车牌作为键值。在填充valid数组时，可以顺便计算该车的停留时间，累加并存在map中。设置maxTime，存储时间最大者。最后，在输出时，遍历map，找出停留时间等于maxTime的即可。

2、map用法：

#include<map>
using namespace std;

map<string, int> mp; //char数组不能作为键

if(mp.count("key_value") == 0)   //查询是否存在
    mp["key_value"] = 1234;

mp["key_value"] = 2345; // 赋值

map<string, int> :: iterator it;     //遍历
for(it = mp.begin(); it != mp.end(); it++){
      it->second = 1111;   // 访问值      
      it->first = "aklsdj";   //访问键  
}

3、C++中使用string 需要include<string>

      字符数组转string： string str;     char c[10] = {'a', 'b', 'c', '\0'};      str = c;

      string转字符数组： string str = “abc”;     char c[10] ;      strcpy(c, str.c_str());

4、将hh:mm:ss转换为秒，可以减少复杂度。

5、注意一辆车反复出入的情况，这时它的总停留时间需要累加。