1153 Decode Registration Card of PAT （25 point(s)）

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 Bfor 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

本文参考了大佬：https://blog.csdn.net/Hickey_Chen/article/details/84933814 

题目大意：N个学生考号，M个问题。题目读懂了也不难，此处就不翻译了，总结一下我遇到的问题吧：开始我把考号中的每一项都写进结构体里，以为处理起来方便点，超时了，之后放一起多了4分。之后便是最坑的cin，cout所需要的时间远超过printf和scanf。改了后19。最后一点就是我将unordered_map定义在了全局变量。。。。。写在里边后满分了。最后感谢大佬代码参考。代码如下：

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <unordered_map>
#include <algorithm>
using namespace std;
const int maxn=1e4+10;
struct card{
	string id;
	int score;
}St[maxn],co[maxn];
struct cr{
	int cnt;
	string tnum;
}Cr[maxn];
bool cmp(card a, card b)
{
	return (a.score==b.score?a.id<b.id:a.score>b.score);
}
bool cmmp(cr a, cr b)
{
	return (a.cnt==b.cnt?a.tnum<b.tnum:a.cnt>b.cnt);
}
int main()
{
//	freopen("1153.txt","r",stdin);
	int n,m,sc,type;
	string str,s;
	cin>>n>>m;
	for(int i=0;i<n;i++){
		cin>>str>>sc;
		St[i].id=str;
		St[i].score=sc;
	}
	for(int i=1;i<=m;i++){
		cin>>type>>s;
		int found=0;
		switch(type){
			case 1:
			{
				int k=0;
				for(int j=0;j<n;j++){
					char L=St[j].id[0];
					if(L==s[0]){                //将符合条件的拷贝出去
						co[k].id=St[j].id;
						co[k].score=St[j].score;
						k++;
						found=1;
					}
				}
				sort(co,co+k,cmp);
//				cout<<"Case "<<i<<": "<<type<<" "<<s<<endl;
				printf("Case %d: %d %s\n",i,type,s.c_str());
				if(!found)	printf("NA\n");//cout<<"NA\n";
				else
					for(int j=0;j<k;j++)
//						cout<<co[j].id<<" "<<co[j].score<<endl;
						printf("%s %d\n",co[j].id.c_str(),co[j].score);
				break;
			}
			case 2:
			{
				int cnt=0,sum=0;
				for(int j=0;j<n;j++){
//					string s1=St[j].testnum;
					string s1=St[j].id.substr(1,3);
					if(s1==s){
						cnt++;
						sum+=St[j].score;
						found=1;
					}
				}
//				cout<<"Case "<<i<<": "<<type<<" "<<s<<endl;
				printf("Case %d: %d %s\n",i,type,s.c_str());
				if(!found)	printf("NA\n");//cout<<"NA\n";
				else
//					cout<<cnt<<" "<<sum<<endl;
					printf("%d %d\n",cnt,sum);
				break;
			}
			case 3:
			{
				int z=0;
				unordered_map<string,int> mp;
				for(int j=0;j<n;j++){
					string s1=St[j].id.substr(4,6);
					string num1=St[j].id.substr(1,3); 
					if(s1==s){
						mp[num1]++;
						found=1;
					}
				}
//				cout<<"Case "<<i<<": "<<type<<" "<<s<<endl;
				printf("Case %d: %d %s\n",i,type,s.c_str());
				if(!found)	printf("NA\n");//cout<<"NA\n";
				else{
					for(auto it=mp.begin();it!=mp.end();it++){
					string key=it->first;
					int value=it->second;	
					Cr[z].tnum=key;                //将结果存放到新的结构体中进行处理
					Cr[z].cnt=value;
					z++;
					}
					sort(Cr,Cr+z,cmmp);
					for(int p=0;p<z;p++){
					//	cout<<Cr[p].tnum<<" "<<Cr[p].cnt<<endl;
						printf("%s %d\n",Cr[p].tnum.c_str(),Cr[p].cnt);
					}	
				}		
				break;
			}
			default:
			//	cout<<"Case "<<i<<": "<<type<<" "<<s<<endl<<"NA\n";
				printf("Case %d: %d %s\nNA\n",i,type,s.c_str());
		}
	}
	return 0;
}