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

日期: 2017年9月7日 23:08 作者: 胡城北 分类: PAT甲级 标签: PAT 阅读(1184) 评论(0)

原题目：

原题链接：https://www.patest.cn/contests/pat-a-practise/1095

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

题目大意

给出车辆出入记录，判断某一时刻停车的数量以及一天内停车时长最多的车辆。 首行N，K表示N条记录，K次询问。 后N行为车辆出入记录。 再K行为询问时间。 输出每个时间的停车数，以及一天内停车时长最多的车辆车牌，若不唯一，按车牌顺序输出。 注意： 给出N条车辆记录，但有用的却不一定有N条，唯有配对成功（in和out）才视为有效记录。比如若出现in in out out情况，第一个in和第二个out因无法配对不视为有效记录。

解题报告

先按第一关键字车牌，第二关键字时间排序，筛选出有效车辆出入记录。 再对有效的进行按时间排序。

代码

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#include "iostream"
#include "algorithm"
#include "vector"
#include "string"
#include "map"
using namespace std;

struct node{
    string car;
    int time;
    string status;
};

bool cmp(struct node a,struct node b){
    return a.time < b.time;
}
bool cmpWithCarNum(struct node a,struct node b){
    if(a.car != b.car)
        return a.car < b.car;
    return a.time < b.time;
}

int N,K;
int maxTime = 0;
map<string,int> parkInTime;
map<string,bool> parkStatus;
vector<struct node> records;
vector<struct node> realRecords;
map<string,int> parkTime;

void init(){
    scanf("%d %d",&N,&K);
    int i;
    string car,status;
    int h,m,s;
    char c1[10];
    char c2[10];
    for(i = 0; i < N; i ++){
        scanf("%s %d:%d:%d %s",c1,&h,&m,&s,c2);
        car = string(c1);
        status = string(c2);
        struct node temp;
        temp.car = car;
        temp.time = h * 3600 + m * 60 + s;
        temp.status = status;
        records.push_back(temp);
    }
}

void cal(){
    parkStatus.clear();
    int i,j;
    int h,m,s,time;
    char c;
    int carNum = 0;
    for(i = 0,j = 0; i < K; i++){
        scanf("%d:%d:%d",&h,&m,&s);
        time = h * 3600 + m * 60 + s;
        while(j < realRecords.size() && realRecords[j].time <= time){
            if(realRecords[j].status == "in"){
                parkStatus[realRecords[j].car] = true;
                parkInTime[realRecords[j].car] = realRecords[j].time;
                carNum ++;
            }else if(parkStatus[realRecords[j].car]){
                parkStatus[realRecords[j].car] = false;
                parkTime[realRecords[j].car] += realRecords[j].time - parkInTime[realRecords[j].car];
                maxTime = max(maxTime,parkTime[realRecords[j].car]);
                carNum --;
            }
            j ++;
        }
        printf("%d\n",carNum);
    }
    while(j < realRecords.size()){
        if(realRecords[j].status == "in"){
            parkStatus[realRecords[j].car] = true;
            parkInTime[realRecords[j].car] = realRecords[j].time;
        }else if(parkStatus[realRecords[j].car]){
            parkStatus[realRecords[j].car] = false;
            parkTime[realRecords[j].car] += realRecords[j].time - parkInTime[realRecords[j].car];
            maxTime = max(maxTime,parkTime[realRecords[j].car]);
        }
        j ++;
    }
}

void fil(){
    for (int i = 0; i < records.size()-1; i++){
        if(records[i].car != records[i+1].car){
            continue;
        }
        if(records[i].status == "in" && records[i+1].status == "out"){
            realRecords.push_back(records[i ++]);
            realRecords.push_back(records[i]);
        }
        continue;
    }
}

int main(){
    init();
    sort(records.begin(),records.end(),cmpWithCarNum);
    fil();
    sort(realRecords.begin(),realRecords.end(),cmp);
    cal();
    for(auto it = parkTime.begin();it != parkTime.end();it++){
        if(maxTime == it->second)
            printf("%s ",it->first.c_str());
    }
    printf("%02d:%02d:%02d\n",maxTime/3600,maxTime%3600/60,maxTime%60);
    system("pause");

}
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