前言

今天不开心就撸了一道PAT的题。所有的合集相关源码我都更新在gitee上了需要自取xingleigao/study - Gitee.com

题目描述

1095 Cars on Campus (30 分)https://pintia.cn/problem-sets/994805342720868352/problems/994805371602845696Zhejiang University has 8 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.

输入描述

Each input file contains one test case. Each case starts with two positive integers N (≤104), the number of records, and K (≤8×104) 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.

输出描述

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.

输入示例

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

输出示例

1
4
5
2
1
0
1
JH007BD ZD00001 07:20:09

解题思路

题目理解

PAT A级别的所有题目都是英文，其实题主作为一个六级都没考过的人都能看懂，相信大家都能看懂吧，看不懂也没关系，经常出现的单词也不多，多看就好了.

停车问题，就是给你车辆的进出记录，然后需要你根据要求输出相应时间点的车辆数目。然后输出停车时间最长的车辆和时间。

有个坑点是in和out交替出现如果同一辆车多个in出现按照最后一个出现的时间点算，其它无效。

完整代码实现

#include<cstdio>
#include<cstring>
#include<string>
#include<map>
#include<algorithm>
using namespace std;
const int maxn = 10010;
struct Car{char id[8];int time;char status[4];
}all[maxn],valid[maxn];
int  num = 0;
map<string, int> partTime;
int timeToInt(int hh,int mm,int ss){return hh * 3600 + mm * 60 + ss;//使用s做单位计时
}
bool cmp(Car a, Car b){ //按照id、时间排序；if(strcmp(a.id,b.id))    return strcmp(a.id,b.id) < 0;else return a.time < b.time;
}
bool cmptime(Car a, Car b){ //按照时间排序；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){      //读入记录scanf("%s%d:%d:%d%s",all[i].id,&hh,&mm,&ss,all[i].status);all[i].time = timeToInt(hh,mm,ss);}sort(all, all + n,cmp);//排序int maxTime = -1;//最长停留时间for(int i = 1;i < n;++i){if(!strcmp(all[i].id,all[i - 1].id) && !strcmp(all[i - 1].status,"in") && !strcmp(all[i].status,"out")){//是否满足条件valid[num++] = all[i - 1];valid[num++] = all[i];int inTime = all[i].time - all[i - 1].time;if(partTime.count(all[i].id) == 0)  partTime[all[i].id] = 0;    //无记录插入记录partTime[all[i].id] += inTime;                                  //计算总停车时间if(partTime[all[i].id] > maxTime)   maxTime = partTime[all[i].id]; //记录最大停留时间}}sort(valid,valid + num,cmptime);int now = 0,numCar = 0;for(int i =0;i < k;++i){                //输出当前时刻的车数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++;else   numCar--;now ++;}printf("%d\n",numCar);}map<string, int>::iterator it;for(it = partTime.begin();it != partTime.end();it ++){if(it->second == maxTime)   printf("%s ",it->first.c_str());}printf("%02d:%02d:%02d\n",maxTime/3600,maxTime%3600/60,maxTime%60);return 0;
}

结果分析与总结

PAT官网所有测试点都通过了。

算法笔记这个题给的有点小坑啊，map完全没学就给了这么一道题，简直离谱。不过c++里面的这些好用的东西确实要好好看看了。

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前言