PAT 甲级 1095 Cars on Campus

1095 Cars on Campus （30 point(s)）

Zhejiang 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.

Input Specification:

Each input file contains one test case. Each case starts with two positive integers N (≤10​4​​), the number of records, and K (≤8×10​4​​) 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

经验总结：

emmmm  这一题，思路走错了，最后一个测试点没通过= =，我是在找到两个匹配的时间（设为start,end），开一个num数组，然后将start到end期间的num全部加一，这就没有充分利用题目所给的条件，就是查询时间是递增的，我是基于查询只查询到固有数据，不需要再进行计算以防超时的思想才使用这种方法，而这一题，就需要在查询中进行计算，建立一个指针一直往后走，利用一个变量实时显示当前时间停车场的车辆数量，然后按照要求输出，这样就不会超时了，这一题的超时和cin，cout没有太大关系，不过还是尽量不要用，这一题也是让我头一次知道，字符数组虽然不能做map中的键，但是可以直接作为string类型为键的map的下标进行访问~，即map[char * ]在map<string , .....> 情况下是可行的~就这些啦

AC代码

#include <cstdio>
#include <vector>
#include <map>
#include <string>
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn=90000;
int n,k,num[maxn]={0},vnum=0;
struct node
{
	int time;
	bool f;
	char id[8];
}Node[10010],valid[10010];
int convert(int h,int m,int s)
{
	return h*3600+m*60+s;
}
bool cmp(node a,node b)
{
	if(strcmp(a.id,b.id))
		return strcmp(a.id,b.id)<0;
	return a.time<b.time;
}
bool cmp1(node a,node b)
{
	return a.time<b.time;
}
int main()
{
	scanf("%d%d",&n,&k);
	char temp[8],status[5];
	int h,m,s,t;
	map<string,vector<node> > mp;
	map<string,int> total;
	for(int i=0;i<n;++i)
	{
		scanf("%s %d:%d:%d",Node[i].id,&h,&m,&s);
		Node[i].time=convert(h,m,s);
		scanf("%s",status);
		if(strcmp(status,"in")==0)
			Node[i].f=0;
		else
			Node[i].f=1;
	}
	int maxtime=-1;
	sort(Node,Node+n,cmp);
	for(int i=0;i<n-1;++i)
	{
		if(!strcmp(Node[i].id,Node[i+1].id)&&Node[i].f==0&&Node[i+1].f==1)
		{
			int start=Node[i].time;
			int end=Node[i+1].time;
			valid[vnum++]=Node[i];
			valid[vnum++]=Node[i+1];
			if(total.count(Node[i].id)==0)
				total[Node[i].id]=end-start;
			else
				total[Node[i].id]+=end-start;
			maxtime=max(total[Node[i].id],maxtime);
		}
	}
	sort(valid,valid+vnum,cmp1);
	int now=0,cnum=0;
	for(int i=0;i<k;++i)
	{
		scanf("%d:%d:%d",&h,&m,&s);
		int time=convert(h,m,s);
		while(now<vnum&&valid[now].time<=time)
		{
			if(valid[now].f==0)
				++cnum;
			else
				--cnum;
			++now;
		}
		printf("%d\n",cnum);
	}
	for(map<string,int>::iterator it=total.begin();it!=total.end();++it)
	{
		if(it->second==maxtime)
			cout<<it->first<<" ";
	}
	printf("%02d:%02d:%02d\n",maxtime/3600,(maxtime%3600)/60,maxtime%60);
	return 0;
}