1095. Cars on Campus (30) <结构体排序>

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 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

给你一些进出车辆的信息，然后给出一个时间，在这个时间的时候校园内有多少辆车

最后输出停车时间最长的车辆

要注意的是车辆有可能车辆的in和out跟时间不对应

首先根据车辆排序 in的车辆flag=1，out车辆flag=-1

if(strcmp(l[i].name,l[i+1].name)==0&&l[i].flag==1&&l[i+1].flag==-1)

说明满足一个车辆进出，把这个车辆的信息保存到新的car结构体中

再根据时间排序

最后输入k个时间的时候，遍历car结构体，sum+=car[start].flag；直接求出总的车辆

if（car[start],time>k） break;

因为输入的时间是递增的，所以start可以记录下来，避免重复遍历数组

最长的时间就用map记录下来，最后输出

全程把时间变成秒来处理

#include<cstdio>#include<cmath>#include<algorithm>#include<iostream>#include<cstring>#include<queue>#include<vector>#include<set>#include<map>#include<stack>using namespace std;typedef struct list{char name[10];int flag=0;int time;}list;bool cmp(list l1,list l2){if(strcmp(l1.name,l2.name)!=0){return strcmp(l1.name,l2.name)<0;}return l1.time<l2.time;}bool cmp2(list l1,list l2){return l1.time<l2.time;}int main(){int n,m;cin>>n>>m;list l[10001];for(int i=0;i<n;i++){scanf("%s",l[i].name);int h,m,s;char fz[4];scanf("%d:%d:%d %s",&h,&m,&s,fz);int ti=h*3600+m*60+s;l[i].time=ti;l[i].flag=strcmp(fz,"in")==0?1:-1;}sort(l,l+n,cmp);/*for(int i=0;i<n;i++){cout<<l[i].name<<" "<<l[i].time<<" "<<l[i].flag<<endl;}*/list car[10001];int cnt=0;int max_time=0;map<string,int> ma;for(int i=0;i<n;i++){if(strcmp(l[i].name,l[i+1].name)==0&&l[i].flag==1&&l[i+1].flag==-1){ma[l[i].name]+=l[i+1].time-l[i].time;if(ma[l[i].name]>max_time) max_time=ma[l[i].name];car[cnt++]=l[i];car[cnt++]=l[i+1];}}sort(car,car+cnt,cmp2);/*for(int i=0;i<cnt;i++){cout<<car[i].name<<" "<<car[i].time<<" "<<car[i].flag<<endl;}*/ int start=0;int sum=0;for(int i=0;i<m;i++){int h,m,s;scanf("%d:%d:%d",&h,&m,&s);int ti=h*3600+m*60+s;for(;start<cnt;start++){if(car[start].time<=ti){sum+=car[start].flag;}else break;}cout<<sum<<endl;}for(map<string,int>::iterator it=ma.begin();it!=ma.end();it++){if(it->second==max_time){printf("%s ",it->first.c_str());}}printf("%02d:%02d:%02d",max_time/3600,max_time%3600/60,max_time%60);return 0;}