1153 Decode Registration Card of PAT (25 分)（模拟，排序，map）

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

题目大意：

给出一组学生的准考证号和成绩，准考证号包含了等级(乙甲顶)，考场号，日期，和个人编号信息，并有三种查询方式
查询一：给出考试等级，找出该等级的考生，按照成绩降序，准考证升序排序
查询二：给出考场号，统计该考场的考生数量和总得分
查询三：给出考试日期，查询改日期下所有考场的考试人数，按照人数降序，考场号升序排序

分析：

先把所有考生的准考证和分数记录下来～
1.按照等级查询，枚举选取匹配的学生，然后排序即可
2.按照考场查询，枚举选取匹配的学生，然后计数、求和
3.按日期查询每个考场人数，用unordered_map存储，最后排序汇总～
注意:1.第三个用map存储会超时，用unordered_map就不会超时啦～
2.排序传参建议用引用传参，这样更快，虽然有时候不用引用传参也能通过，但还是尽量用，养成好习惯～

原文链接：https://blog.csdn.net/liuchuo/article/details/84973049

题解

这题在做乙级的时候很顺啊Orz乙级原题题解

#include <bits/stdc++.h>

using namespace std;
struct node
{
    string id;
    int score;
};
bool cmp(node a,node b){
    if(a.score!=b.score) return a.score>b.score;
    else return a.id<b.id;
}
int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("1.txt", "r", stdin);
#endif
    int n,m,k,cnt;
    string s;
    scanf("%d%d",&n,&m);
    vector<node> v(n);
    for(int i=0;i<n;i++){
        cin>>v[i].id>>v[i].score;
    }
    for(int i=0;i<m;i++){
        cin>>k>>s;
        printf("Case %d: %d %s\n",i+1,k,s.c_str());
	//忘记清空的后果是只能过1个样例
        vector<node> tmp,result;
        if(k==1){
            for(int j=0;j<n;j++){
                if(v[j].id[0]==s[0]){
                    tmp.push_back(v[j]);
                }
            }
            if(tmp.size()!=0){
                sort(tmp.begin(),tmp.end(),cmp);
                for(int j=0;j<tmp.size();j++)
                    printf("%s %d\n",tmp[j].id.c_str(),tmp[j].score);
            }
        }else if(k==2){
            int sum=0;
            cnt=0;
            for(int j=0;j<n;j++){
                if(v[j].id.substr(1,3)==s){
                    cnt++;
                    sum+=v[j].score;
                }
            }
            if(cnt!=0) cout<<cnt<<" "<<sum<<endl;
        }else if(k==3){
            unordered_map<string,int> mp;
            for(int j=0;j<n;j++){
                if(v[j].id.substr(4,6)==s){
                    mp[v[j].id.substr(1,3)]++;
                }
            }
            for(auto it=mp.begin();it!=mp.end();it++){
                result.push_back({it->first,it->second});
            }
            if(result.size()!=0){
                sort(result.begin(),result.end(),cmp);
                for(int j=0;j<result.size();j++)
                    printf("%s %d\n",result[j].id.c_str(),result[j].score);
            }
        }
        if((k==1&&tmp.size()==0)||(k==2&&cnt==0)||(k==3&&result.size()==0))
            printf("NA\n");
    }
    return 0;
}