1095 解码PAT准考证 (25 point(s))

• 写的时候有个几问题，刚开始以为考场的数据能够和日期的相似而共用，但是测试的时候发现，同考场的不一定同日期。所以又新建了一个结构体来存日期相关的数据。

• 测试点一、二、四。

  题干太长看得有点意识模糊。写的时候只考虑了类型 2 查询无果，而没有考虑类型 1 和 3 。所以要参考类型 2 一样多加一个查询判断。

  我是用下面这个测试数据才发现自己忽略了这个判断的。

  测试点一、二、四

• 测试点三。修改了上面的问题后剩下测试点三超时。又参考了其他文章，原来是老问题 cin cout 太慢了，改成 printf 和 scanf 即可。

#include <bits/stdc++.h>
using namespace std;

struct Stu{
	string id;
	int score;
	
	bool operator < (Stu s) const{
		if(score != s.score) return score > s.score;
		return id < s.id;
	}
};

struct Room{
	string id;
	int score = 0, peo = 0;
	
	bool operator < (Room r) const{
		if(peo != r.peo) return peo > r.peo;
		return id < r.id;
	}
};

struct Date{
	string id;
	int peo = 0;
	
	bool operator < (Room r) const{
		if(peo != r.peo) return peo > r.peo;
		return id < r.id;
	}
};

int main() {
	int N, M;
	map<char, set<Stu>> stu;
	map<string, Room> room;
	map<string, map<string, Room>> date;
	cin >> N >> M;
	
	while(N--){
		string id;
		int score;
		cin >> id;
		scanf("%d", &score);
		string roomID =  id.substr(1, 3), datetmp =  id.substr(4, 6);

		// 类型1 
		stu[id[0]].insert({id, score});	
		
		// 类型2
		room[roomID].id = roomID;
		room[roomID].score += score;
		room[roomID].peo++;
		
		// 类型3 
		date[datetmp][roomID].id = roomID;
		date[datetmp][roomID].peo++;
	}
	
	for(int i = 1; i <= M; i++){
		int type; string obj; 
		cin >> type >> obj;
		printf("Case %d: %d %s\n", i, type, obj.c_str());
		
		if(type == 1){
			if(stu.find(obj[0]) == stu.end())cout << "NA" << endl;
			else 
				for(auto s: stu[obj[0]])
					printf("%s %d\n", s.id.c_str(), s.score);
		}
		else if(type == 2){
			if(room.find(obj) == room.end())cout << "NA" << endl;
			else printf("%d %d\n", room[obj].peo, room[obj].score);
		}
		else{
			if(date.find(obj) == date.end())cout << "NA" << endl;
			else{
				// 放入 set 集合排序 
				set<Room> out;
				for(auto d: date[obj]) out.insert(d.second);
				for(auto o: out)
					printf("%s %d\n", o.id.c_str(), o.peo);
			} 
			
		}
	}
}

// 15 points
#include <bits/stdc++.h>
using namespace std;

struct Stu{
	string id;
	int score;
	
	bool operator < (Stu s) const{
		if(score != s.score) return score > s.score;
		return id < s.id;
	}
};

struct Room{
	string id;
	int score = 0, peo = 0;
	
	bool operator < (Room r) const{
		if(peo != r.peo) return peo > r.peo;
		return id < r.id;
	}
};

struct Date{
	string id;
	int peo = 0;
	
	bool operator < (Room r) const{
		if(peo != r.peo) return peo > r.peo;
		return id < r.id;
	}
};

int main() {
	int N, M;
	map<char, set<Stu>> stu;
	map<string, Room> room;
	map<string, map<string, Room>> date;
	cin >> N >> M;
	
	while(N--){
		string id;
		int score;
		cin >> id >> score;
		string roomID =  id.substr(1, 3), datetmp =  id.substr(4, 6);

		// 类型1 
		stu[id[0]].insert({id, score});	
		
		// 类型2
		room[roomID].id = roomID;
		room[roomID].score += score;
		room[roomID].peo++;
		
		// 类型3 
		date[datetmp][roomID].id = roomID;
		date[datetmp][roomID].peo++;
	}
	
	for(int i = 1; i <= M; i++){
		int type; string obj; 
		cin >> type >> obj;
		printf("Case %d: %d %s\n", i, type, obj.c_str());
		
		if(type == 1){
			for(auto s: stu[obj[0]])
				cout << s.id << " " << s.score << endl;
		}
		else if(type == 2){
			if(room.find(obj) == room.end())cout << "NA" << endl;
			else cout << room[obj].peo << " " << room[obj].score << endl;
		}
		else{
			// 放入 set 集合排序 
			set<Room> out;
			for(auto d: date[obj]) out.insert(d.second);
			for(auto o: out)
				cout << o.id << " " << o.peo << endl;
		}
	}
}

标签：1095,PAT,score,point,int,peo,return,roomID,id
来源： https://www.cnblogs.com/Atl212/p/15321281.html