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最新推荐文章于 2023-05-08 23:27:18 发布

原创最新推荐文章于 2023-05-08 23:27:18 发布 · 301 阅读

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#模拟事件 #PAT

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本文介绍了一种处理有序时间记录的方法，通过将记录排序并遍历一次来高效获取所需结果。文章提供了一个完整的C++实现示例，展示了如何处理进出记录，并统计特定时间段内的人数及最长时间段的人员名单。

卡时间

再者题目中强调请求时间有序

所以把记录按时间排好序，然后按序访问，这样只要访问一遍就可以得到所有的结果

代码如下：

#include<iostream>
#include<vector>
#include<algorithm>
#include<string>
#include<map>
using namespace std;
struct recode{
	string pname;
	//int hh, mm, ss;
	int t;
	bool in;//为in即为true，out为false 
};
int n, k;
vector<recode> all_recodes,right_recodes;
map<string,int> mp;
int changetime(int hh, int mm, int ss){
	return hh*3600+mm*60+ss;
}
bool comparetime(recode a, recode b){
	if(a.t < b.t) return true;
	else return false;
}
bool compare(recode a, recode b){
	if(a.pname < b.pname) return true;
	else if(a.pname == b.pname && a.t < b.t) return true;
	else return false;
} 
int main(){
	cin>>n>>k;
	for(int i = 0; i < n; i++){
		recode a;
		cin>>a.pname;
		int hh, mm, ss;
		scanf("%d:%d:%d",&hh,&mm,&ss);
		a.t=changetime(hh,mm,ss);
		string s;
		cin>>s;
		if(s == "in"){
			a.in = true;
		}
		else{
			a.in = false;
		}
		all_recodes.push_back(a);
	}
	sort(all_recodes.begin(),all_recodes.end(),compare);
//	printf("\n");
//	for(int i = 0; i < all_recodes.size();i++){
//		printf("%s %d:%d:%d %d\n",all_recodes[i].pname.c_str(),all_recodes[i].t/3600,all_recodes[i].t%3600/60,all_recodes[i].t%3600%60,all_recodes[i].in);
//	
//	}
	for(int i = 0; i < all_recodes.size()-1; i++){
		if(all_recodes[i].pname == all_recodes[i+1].pname && all_recodes[i].in == true && all_recodes[i+1].in == false){
			right_recodes.push_back(all_recodes[i]);
			right_recodes.push_back(all_recodes[i+1]);
			mp[all_recodes[i].pname] += all_recodes[i+1].t - all_recodes[i].t;
//		    printf("%s %02d:%02d:%02d in\n",all_recodes[i].pname.c_str(),all_recodes[i].t/3600,all_recodes[i].t%3600/60,all_recodes[i].t%3600%60);
//			printf("%s %02d:%02d:%02d out\n",all_recodes[i+1].pname.c_str(),all_recodes[i+1].t/3600,all_recodes[i+1].t%3600/60,all_recodes[i+1].t%3600%60);
			i++;
		}	
	}
	sort(right_recodes.begin(),right_recodes.end(),comparetime);
	int count = 0;
	int j = 0;
	for(int i = 0; i < k; i++){
		int hh, mm, ss;
		scanf("%d:%d:%d",&hh,&mm,&ss);
		int qt = changetime(hh, mm, ss);
		while(right_recodes[j].t <= qt && j<right_recodes.size()){
			if(right_recodes[j].in == true) count++;
			else count--;
			j++;
		}
		printf("%d\n",count);
	}
	int maxt = 0;
	vector<string> max;
	for(map<string,int>::iterator it = mp.begin(); it != mp.end(); it++){
		if(it->second > maxt){
			maxt = it->second;
			max.clear();
			max.push_back(it->first);
		}
		else if(maxt == it->second){
			max.push_back(it->first);
		}
	}
	for(int i = 0; i < max.size(); i++){
		printf("%s ",max[i].c_str());
	}
	printf("%02d:%02d:%02d",maxt/3600,maxt%3600/60,maxt%3600%60);
	return 0;
}