PAT_A1153#Decode Registration Card of PAT

Source:

PAT A1153 Decode Registration Card of PAT (25 分)

Description:

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 (≤) and M (≤), the numbers of cards and the queries, respectively.

Then N lines follow, each gives a card number and the owner's score (integer in [), 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

Keys:

• map（C++ STL）
• string（C++ STL）
• 快乐模拟

Attention:

• 每轮mp要记得清空-,-
• cmp引用传参效率更快
• cout 输出会超时

Code:

 /*
 Data: 2019-08-02 21:39:05
 Problem: PAT_A1153#Decode Registration Card of PAT
 AC: 34:30

 题目大意：
 第一行：给出卡片数量n<=1e4，和查询数量m<=100
 接下来n行，给出卡号，成绩[0,100]
 接下来m行，给出查询
 1. 给出指定等级，输出各个考生，及其成绩，递减，卡号递增
 2. 给出指定地点，输出考生总数，及其总分，
 3. 给出指定时间，输出各个考场的考生总数，人数递减，地点递增
 4. 查询失败输出NA
 */
 #include<cstdio>
 #include<map>
 #include<string>
 #include<vector>
 #include<iostream>
 #include<algorithm>
 using namespace std;
 const int M=1e4+;
 struct node
 {
     string id;
     int score;
 }info[M];

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

 void Query(int k, int n)
 {
     int index,sum=,cnt=;
     string s;
     cin >> index >> s;
     printf("Case %d: %d %s\n", k,index,s.c_str());
     vector<node> ans;
     map<string,int> mp;
     for(int i=; i<n; i++)
     {
         if(index== && info[i].id[]==s[])
             ans.push_back(info[i]);
         else if(index== && info[i].id.substr(,)==s)
         {
             cnt++;
             sum += info[i].score;
         }
         else if(index== && info[i].id.substr(,)==s)
             mp[info[i].id.substr(,)]++;
     }
     if(index== && cnt!=)
         printf("%d %d\n", cnt, sum);
     else
         for(auto it=mp.begin(); it!=mp.end(); it++)
             ans.push_back({it->first,it->second});
     sort(ans.begin(),ans.end(),cmp);
     for(int i=; i<ans.size(); i++)
         printf("%s %d\n", ans[i].id.c_str(),ans[i].score);
     if(ans.size()== && cnt==)
         printf("NA\n");

     return;
 }

 int main()
 {
 #ifdef ONLINE_JUDGE
 #else
     freopen("Test.txt", "r", stdin);
 #endif // ONLINE_JUDGE

     int n,m;
     scanf("%d%d", &n,&m);
     for(int i=; i<n; i++)
         cin >> info[i].id >> info[i].score;
     for(int i=; i<=m; i++)
         Query(i,n);

     return ;
 }