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 (≤) 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

 

解码PAT准考证，我们已知PAT准考证，第一位是TAB（代表基本），二到四是代表考场，五到十位代表日期，最后三位代表考场人数。

我们要进行查询，如果输入1，则查询输入考试级别的分数降序，名字升序排序。如果输入2，则查询该考场人数和总分，输入3，则查询该日期的考场和考场中的人数。

#include <iostream>
#include <vector>
#include <unordered_map>
#include <algorithm>
using namespace std;
int N, M, choose;
string tmp;
pair<string, int> cards[11000];
bool cmp(pair<string, int> p1, pair<string, int> p2) {
    return p1.second != p2.second ? p1.second > p2.second : p1.first < p2.first;
}
int main() {
    cin >> N >> M;
    for(int i = 0; i < N; i++)  
        cin >> cards[i].first >> cards[i].second;
    for(int c = 1; c <= M; c++) {
        cin >> choose >> tmp;
        printf("Case %d: %d %s\n", c, choose, tmp.c_str());
        if(choose == 1) {
            vector<pair<string, int>> v;
            for(int i = 0; i < N; i++) 
                if(cards[i].first[0] == tmp[0]) v.push_back(cards[i]);
            if(v.size() != 0) {
                sort(v.begin(), v.end(), cmp);
                for(auto x : v) printf("%s %d\n", x.first.c_str(), x.second);
            } else printf("NA\n");
        } else if(choose == 2) {
            int cnt = 0, all_score = 0;
            for(int i = 0; i < N; i++) {
                if(cards[i].first.substr(1, 3) == tmp) {
                    cnt++; 
                    all_score += cards[i].second;
                }
            }
            if(cnt != 0) printf("%d %d\n", cnt, all_score);
            else printf("NA\n");
        }else {
            unordered_map<string, int> m;
            for(int i = 0; i < N; i++) 
                if(cards[i].first.substr(4, 6) == tmp) 
                    m[cards[i].first.substr(1, 3)]++;
            if(m.size() != 0) {
                vector<pair<string, int>> v(m.begin(), m.end());
                sort(v.begin(), v.end(), cmp);
                for(auto x: v) printf("%s %d\n", x.first.c_str(), x.second);
            } else printf("NA\n");
        }
    }
    return 0;
}
// B 123 180908 127
// 1 decreasing
// 2 Nt Ns
// 3 Nnumber testee