A1095 Cars on Campus [排序]

题目关键:给时间算出该时间在校内车辆
最后一行写出停留时间最长的车子的名字和和停留时间
关键点：map容器使用，根据第一个元素排序
还有这个Note: the queries are given in ascending order of the times.
查询时间从小到大我草，顺着查一次就行了，不然有三个会超时，坑点真的太多了
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#include<iostream>
#include<algorithm>
#include<cstring>
#include<string>
#include<map>
#include<cstdio>
const int maxn = 10010;
using namespace std;
struct car
{
	char id[15];
	int time;
	char status[4];
}all[maxn], valid[maxn];
int num = 0;
map<string, int>_time;
bool cmp1(car a, car b)
{
	if (strcmp(a.id, b.id))
		return strcmp(a.id, b.id) < 0;
	else
		return a.time < b.time;
}
bool cmp2(car a, car b)
{
	return a.time < b.time;
}
int timetoint(int hh, int mm, int ss)
{
	return hh * 3600 + mm * 60 + ss;
}
int main()
{
	int n, k, hh, mm, ss;
	scanf("%d%d", &n, &k);
	for (int i = 0; i < n; i++)
	{
		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, cmp1);
	int maxtime = -1;
	for (int i = 0; i < n - 1; i++)
	{
		if (!strcmp(all[i].id, all[i + 1].id) && !strcmp(all[i].status, "in") && !strcmp(all[i + 1].status, "out"))
		{
			valid[num++] = all[i];
			valid[num++] = all[i + 1];
			int intime = all[i + 1].time - all[i].time;
			if (_time.count(all[i].id) == 0)
			{
				_time[all[i].id] = 0;
			}
			_time[all[i].id] += intime;
			maxtime = max(maxtime, _time[all[i].id]);
		}
	}
	//开始对有效数组进行操作了
	sort(valid, valid + num, cmp2);
	int now=0,numcar = 0;
	for (int i = 0; i < k; i++)
	{
		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 = _time.begin(); it != _time.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;
}