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>#include<map>#include<algorithm>#include<iostream>using namespace std;const int maxn=10010;struct Car{    char id[8];//车牌     int time;    char status[4];// }all[maxn],valid[maxn];//有效记录和总记录 int num=0;//有效记录个数 map<string,int> parkTime;//车牌号->停留时间 int timeToInt(int hh,int mm,int ss){//把时间按秒存储，便于操作     return hh*3600+mm*60+ss;} //以下两种不同的排序方法，//一种对总记录进行排序，按车牌和出入时间排序，为了便于找出有效记录 bool cmpByIdAndTime(Car a,Car b){    if(strcmp(a.id,b.id)) return strcmp(a.id,b.id)<0;    else return a.time<b.time;}//一种对有效记录排序 ，按出入时间排序，为了便于找出某个时间点校内的车辆数 bool cmpByTime(Car a,Car b){    return a.time<b.time;}int main(){    int n,k,hh,mm,ss;    scanf("%d%d",&n,&k);    for(int i=0;i<n;i++){        scanf("%s %02d:%02d:%02d %s",all[i].id,&hh,&mm,&ss,all[i].status);        all[i].time=timeToInt(hh,mm,ss);    }    sort(all,all+n,cmpByIdAndTime);//  printf("总记录排序后:\n");//  for(int i=0;i<n;i++){ //      printf("%s %02d:%02d:%02d %s\n",all[i].id,all[i].time/3600,all[i].time%3600/60,all[i].time%60,all[i].status);//  }    int maxTime=-1;    for(int i=0;i<n-1;i++){//把有效记录放入valid数组         if(!strcmp(all[i].id,all[i+1].id)&&!strcmp(all[i].status,"in")&&!strcmp(all[i+1].status,"out")){            valid[num++]=all[i];            valid[num++]=all[i+1];            int inTime=all[i+1].time-all[i].time;            if(parkTime.count(all[i].id)==0){//map中还没有这个车牌号                 parkTime[all[i].id]=0;      //数量置零             }            parkTime[all[i].id]+=inTime;//增加给车牌号的总停留时间             maxTime=max(maxTime,parkTime[all[i].id]);        }    }    sort(valid,valid+num,cmpByTime);//把有效记录按时间从小到大排序//  printf("有效记录排序后:\n");//  for(int i=0;i<num;i++){//      printf("%s %02d:%02d:%02d %s\n",valid[i].id,valid[i].time/3600,valid[i].time%3600/60,valid[i].time%60,valid[i].status);//  }     int now=0,numCar=0;    //以下是本题最巧妙的一步，要求某一个时间点校内人数，    //只需要求出该时间点之前进入的车辆和出去的车辆之差     for(int i=0;i<k;i++){        scanf("%d:%d:%d",&hh,&mm,&ss);        int time=timeToInt(hh,mm,ss);        while(now<num&&valid[now].time<=time){            if(!strcmp(valid[now].status,"in")) numCar++;//入             else numCar--;//出             now++;        }        printf("%d\n",numCar);    }     map<string,int>::iterator it;    for(it=parkTime.begin();it!=parkTime.end();it++){        if(it->second==maxTime){            printf("%s ",it->first.c_str());//这个first.c_str()是将c++的string对象类型转化为c语言中的字符数组类型                                            //若不转换会编译错误：[Error] cannot pass objects of non-trivially-copyable type 'const class std::basic_string<char>//          cout<<it->first<<" ";也可以用这个方法         }    }    printf("%02d:%02d:%02d\n",maxTime/3600,maxTime%3600/60,maxTime%60);    return 0;}/*运行调试 16 7JH007BD 18:00:01 inZD00001 11:30:08 outDB8888A 13:00:00 outZA3Q625 23:59:50 outZA133CH 10:23:00 inZD00001 04:09:59 inJH007BD 05:09:59 inZA3Q625 11:42:01 outJH007BD 05:10:33 inZA3Q625 06:30:50 inJH007BD 12:23:42 outZA3Q625 23:55:00 inJH007BD 12:24:23 outZA133CH 17:11:22 outJH007BD 18:07:01 outDB8888A 06:30:50 in总记录排序后:DB8888A 06:30:50 inDB8888A 13:00:00 outJH007BD 05:09:59 inJH007BD 05:10:33 inJH007BD 12:23:42 outJH007BD 12:24:23 outJH007BD 18:00:01 inJH007BD 18:07:01 outZA133CH 10:23:00 inZA133CH 17:11:22 outZA3Q625 06:30:50 inZA3Q625 11:42:01 outZA3Q625 23:55:00 inZA3Q625 23:59:50 outZD00001 04:09:59 inZD00001 11:30:08 out有效记录排序后:ZD00001 04:09:59 inJH007BD 05:10:33 inDB8888A 06:30:50 inZA3Q625 06:30:50 inZA133CH 10:23:00 inZD00001 11:30:08 outZA3Q625 11:42:01 outJH007BD 12:23:42 outDB8888A 13:00:00 outZA133CH 17:11:22 outJH007BD 18:00:01 inJH007BD 18:07:01 outZA3Q625 23:55:00 inZA3Q625 23:59:50 out05:10:00106:30:50411:00:00512:23:42214:00:00118:00:00023:59:001JH007BD ZD00001 07:20:09*/