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 ascendingorder 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 23:59:00Sample Output:
1 4 5 2 1 0 1 JH007BD ZD00001 07:20:09
#include<cstdio>
#include<iostream>
#include<vector>
#include<algorithm>
#include<queue>
#include<stack>
#include<string>
#include<string.h>
#include<map>
using namespace std;
struct Record{
int time;
bool status;
Record(int t,int s):time(t),status(s){}
};
bool cmp(Record* r1, Record* r2){
return r1->time < r2->time;
}
int Time2int(int h, int m, int s){
return h * 60 * 60 + m * 60 + s;
}
vector<vector<Record*>>V;
map<string, int>M;
map<int, string>M2;
int main(){
freopen("F://Temp/input.txt", "r", stdin);
int record_num, query_num;
cin >> record_num >> query_num;
int M_idx = 0;
for (int i = 0; i < record_num; ++i){
string str;
int hour, min, sec;
string sta;
cin >> str;
scanf("%d:%d:%d", &hour, &min, &sec);
cin >> sta;
Record *rec = new Record(Time2int(hour,min,sec),sta == "in"?true:false);
if (M.find(str) == M.end()){
M2[M_idx] = str;
M[str] = M_idx++;
vector<Record*>VR;
VR.push_back(rec);
V.push_back(VR);
}
else
V[M[str]].push_back(rec);
}
int *sum_time = new int[V.size()];
for (int i = 0; i < V.size(); ++i){
sum_time[i] = 0;
}
vector<Record*>VT;
for (int i = 0; i < V.size(); ++i){
sort(V[i].begin(), V[i].end(), cmp);
int j = 0;
//vector<Record*>V_tmp;
for (int j = 0; j < V[i].size() - 1; j++){
if (V[i][j]->status == true && V[i][j + 1]->status == false){//in和out成对出现
//V_tmp.push_back(V[i][j]);
//V_tmp.push_back(V[i][j + 1]);
VT.push_back(V[i][j]);
VT.push_back(V[i][j + 1]);
sum_time[i] += V[i][j + 1]->time - V[i][j]->time;//顺便把每辆车时间算出来
}
}
//V[i] = V_tmp;//整理出有用的成对的记录
}
//因为查询很多，所以可以把汽车整理出的记录集中排序，然后按照时间来做一个循环
sort(VT.begin(), VT.end(), cmp);
//for (int i = 0; i < VT.size(); ++i){
// int tt = VT[i]->time;
// cout << tt / 3600 << ":" << tt / 60 % 60 << ":" << tt % 60 << " ";
// if (VT[i]->status)
// cout << "in" << endl;
// else
// cout << "out" << endl;
//}
int *car_sum = new int[VT.size()];
car_sum[0] = 1;
for (int i = 1; i < VT.size(); ++i){
if (VT[i]->status == true)car_sum[i] = car_sum[i - 1] + 1;
else car_sum[i] = car_sum[i - 1] - 1;
}
int time_idx = 0;//用来指示上一个查询的时间位置在哪里
for (int i = 0; i < query_num; ++i){
int hour, min, sec;
scanf("%d:%d:%d", &hour, &min, &sec);
int query_t = Time2int(hour, min, sec);
while (time_idx < VT.size() && VT[time_idx]->time <= query_t)time_idx++;
cout << car_sum[time_idx - 1]<< endl;
}
//汽车各自的记录可以找出哪一辆车是停得最长的
int max_time = 0;
vector<int>V_max_idx;
for (int i = 0; i < V.size(); ++i){
if (sum_time[i] > max_time){
V_max_idx.clear();
V_max_idx.push_back(i);
max_time = sum_time[i];
}
else if (sum_time[i] == max_time){
V_max_idx.push_back(i);
}
}
for (int i = 0; i < V_max_idx.size(); ++i){
cout << M2[V_max_idx[i]] << " ";
}
printf("%02d:%02d:%02d", max_time / 3600, max_time / 60 % 60, max_time % 60);
cout << endl;
return 0;
}
