PAT A1153 2019.08.14 【结构体sort map】

1153 Decode Registration Card of PAT (25 分)

A registration card number of PAT consists of 4 parts:

• the 1st letter represents the test level, namely, T for the top level, A for advance and B for basic;
• the 2nd - 4th digits are the test site number, ranged from 101 to 999;
• the 5th - 10th digits give the test date, in the form of yymmdd;
• finally the 11th - 13th digits are the testee's number, ranged from 000 to 999.

Now given a set of registration card numbers and the scores of the card owners, you are supposed to output the various statistics according to the given queries.

Input Specification:

Each input file contains one test case. For each case, the first line gives two positive integers N (≤10​4​​) and M (≤100), the numbers of cards and the queries, respectively.

Then N lines follow, each gives a card number and the owner's score (integer in [0,100]), separated by a space.

After the info of testees, there are M lines, each gives a query in the format Type Term, where

• Type being 1 means to output all the testees on a given level, in non-increasing order of their scores. The corresponding Term will be the letter which specifies the level;
• Type being 2 means to output the total number of testees together with their total scores in a given site. The corresponding Term will then be the site number;
• Type being 3 means to output the total number of testees of every site for a given test date. The corresponding Term will then be the date, given in the same format as in the registration card.

Output Specification:

For each query, first print in a line Case #: input, where # is the index of the query case, starting from 1; and input is a copy of the corresponding input query. Then output as requested:

• for a type 1 query, the output format is the same as in input, that is, CardNumber Score. If there is a tie of the scores, output in increasing alphabetical order of their card numbers (uniqueness of the card numbers is guaranteed);
• for a type 2 query, output in the format Nt Ns where Nt is the total number of testees and Ns is their total score;
• for a type 3 query, output in the format Site Nt where Site is the site number and Nt is the total number of testees at Site. The output must be in non-increasing order of Nt's, or in increasing order of site numbers if there is a tie of Nt.

If the result of a query is empty, simply print NA.

Sample Input:

8 4
B123180908127 99
B102180908003 86
A112180318002 98
T107150310127 62
A107180908108 100
T123180908010 78
B112160918035 88
A107180908021 98
1 A
2 107
3 180908
2 999

Sample Output:

Case 1: 1 A
A107180908108 100
A107180908021 98
A112180318002 98
Case 2: 2 107
3 260
Case 3: 3 180908
107 2
123 2
102 1
Case 4: 2 999
NA

思路分析

做完此题，结构体的排序应该就没问题了

此处用先排序，再输出的方式肯定会超时（19/25）

未超时的版本以后再写

#include<cstdio>
#include<iostream>
#include<map> 
#include<string>
#include<vector>
#include<algorithm>
using namespace std;

typedef struct STU{
	string id;
	char level;
	string site;
	string date;
	string test;
	int score;
}STU;

typedef struct CASE3{
	string site;
	int num;
}CASE3;

STU create(string id,int score)
{
	STU stu;
	stu.id=id;
	stu.level=id[0];
	stu.site=id.substr(1,3);
	stu.date=id.substr(4,6);
	stu.test=id.substr(10,12);
	stu.score=score;
	return stu;
}

vector<STU>v;

bool cmp(STU a,STU b)
{
	if(a.score!=b.score)
	  return a.score>b.score;
	else
	  return a.id<b.id;
}

bool cmp3(CASE3 a,CASE3 b)
{
	if(a.num!=b.num)
	  return a.num>b.num;
	else 
	  return a.site<b.site;
}

void check(vector<STU>v)
{
	for(int i=0;i<v.size();i++)
	{
		cout<<v[i].level<<" ";
		cout<<v[i].site<<" ";
		cout<<v[i].date<<" ";
		cout<<v[i].test<<"  ---";
		cout<<v[i].score<<endl;
	}
}



int main()
{
	int n,m;cin>>n>>m;
	for(int i=0;i<n;i++)
	{
		string id;int score;
		cin>>id>>score;
		STU temp = create(id,score);
		v.push_back(temp);
	}
//	check(v);
	
	for(int i=1;i<=m;i++)
	{
		int pos;cin>>pos;
		if(pos==1)
		{
			bool isNull=1;
			char level;cin>>level;
			cout<<"Case "<<i<<": "<<pos<<" "<<level<<endl;
			sort(v.begin(),v.end(),cmp);
			for(int j=0;j<v.size();j++)
			  if(v[j].level==level)
			  {
			  	cout<<v[j].id<<" "<<v[j].score<<endl;
			  	isNull=0;
			  }
			if(isNull==1)cout<<"NA"<<endl;
		}
		
		if(pos==2)
		{
			string site;cin>>site;
			cout<<"Case "<<i<<": "<<pos<<" "<<site<<endl;
			int total=0;
			int num=0;
			for(int j=0;j<v.size();j++)
			  if(v[j].site==site)
			  {
			  	num++;
			  	total+=v[j].score;
			  }
			if(num==0&&total==0)cout<<"NA"<<endl;
			else cout<<num<<" "<<total<<endl;
		}
		
		if(pos==3)
		{
			string date;cin>>date;
			cout<<"Case "<<i<<": "<<pos<<" "<<date<<endl;
			string stack[10010];int top=-1;
			map<string,int>cnt;
			for(int j=0;j<v.size();j++)
			  if(v[j].date==date)
			  {
			  	if(cnt[v[j].site]==0)stack[++top]=v[j].site;
			  	cnt[v[j].site]++;
			  }
			vector<CASE3>case3;
			for(int j=0;j<=top;j++)
			{
				CASE3 temp;
				temp.site=stack[j];
				temp.num=cnt[stack[j]];
				case3.push_back(temp);
			}
			sort(case3.begin(),case3.end(),cmp3);
			for(int j=0;j<case3.size();j++)
			 cout<<case3[j].site<<" "<<case3[j].num<<endl;
			if(case3.size()==0)
			 cout<<"NA"<<endl;
		}	
		
	}
	
}