PAT A 1095. Cars on Campus (30)

题目

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; andstatus 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 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 in05:10:0006:30:5011:00:0012:23:4214:00:0018:00:0023:59:00

Sample Output:

1452101JH007BD ZD00001 07:20:09

比较肉的模拟类问题，知道怎么去除无效的数据，直接模拟就可以了

有效的配对是指紧挨着的两个in和out，

也就是说连续的多个in取最后一个为有效，连续多个out取第一个为有效，

最后如果有in没有被匹配，则应该放弃相应的记录

代码：

图省事，写得比较烂，实际是不需要对数据进行那么多的复制操作的

#include <iostream>#include <algorithm>#include <string>#include <vector>#include <map>using namespace std;const int LIMIT=24*3600;struct car//车结构{int id,time,in_out;//车的id，动作时间，进出动作car(int i=0,int t=0,int io=0):id(i),time(t),in_out(io){}};int to_time(string s);//将字符串表示的时间转换为时间bool cm(const car &c1,const car &c2);//时间排序int main(){int n,k;cin>>n>>k;map<string,int> name_to_id;//车牌到id转换vector<string> id_to_name(1,"");//id到车牌转换int id=0;vector<car> data;//存储的数据car tempc;//临时输入数据string sn,st,sf;for(int i=0;i<n;i++)//输入所有数据{cin>>sn>>st>>sf;if(name_to_id[sn]==0){name_to_id[sn]=++id;id_to_name.push_back(sn);}tempc.id=name_to_id[sn];tempc.time=to_time(st);if(sf=="in")tempc.in_out=1;elsetempc.in_out=0;data.push_back(tempc);}sort(data.begin(),data.end(),cm);//时间排序vector<vector<car>> use(id+5);//各个id的车的有效记录vector<int> total(id+5,0);//各个id的车的总泊车时间for(int i=0;i<data.size();i++)//按id划分，去除不必要数据{if(data[i].in_out==1){if(!use[data[i].id].empty()&&use[data[i].id].back().in_out==1)use[data[i].id].pop_back();use[data[i].id].push_back(data[i]);}else{if(!use[data[i].id].empty()&&use[data[i].id].back().in_out==1){total[data[i].id]+=data[i].time-use[data[i].id].back().time;use[data[i].id].push_back(data[i]);}}}for(int i=0;i<use.size();i++){if(!use[i].empty()&&use[i].back().in_out==1)use[i].pop_back();}int max=0;//最长泊车时间vector<string> max_name;//最长泊车时间的车辆的车牌集合for(int i=1;i<=id;i++)//统计{if(!use[i].empty()&&total[i]==max)max_name.push_back(id_to_name[use[i].front().id]);else if(!use[i].empty()&&total[i]>max){max=total[i];max_name.clear();max_name.push_back(id_to_name[use[i].front().id]);}}sort(max_name.begin(),max_name.end());data.clear();//清除原数据，重新取所有有效记录，并排序for(int i=1;i<=id;i++)for(int j=0;j<use[i].size();j++)data.push_back(use[i][j]);sort(data.begin(),data.end(),cm);int pos=0,count=0,next;//取到的位置，当前学校中的车辆数，下个请求时间for(int i=0;i<k;i++)//统计输出{cin>>st;next=to_time(st);while(pos<data.size()&&data[pos].time<=next){if(data[pos].in_out==1)count++;elsecount--;pos++;}printf("%d\n",count);}for(int i=0;i<max_name.size();i++)cout<<max_name[i]<<" ";printf("%02d:%02d:%02d",max/3600,max%3600/60,max%60);return 0;}int to_time(string s){int time=0;time+=(s[0]-'0')*10*3600;time+=(s[1]-'0')*3600;time+=(s[3]-'0')*10*60;time+=(s[4]-'0')*60;time+=(s[6]-'0')*10;time+=(s[7]-'0');return time;}bool cm(const car &c1,const car &c2){return c1.time<c2.time;}