PAT A1153 Decode Registration Card of PAT （25 分）——多种情况排序

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

#include <stdio.h>
#include <string>
#include <iostream>
#include <algorithm>
#include <vector>
#include <map>
#include <unordered_map>
using namespace std;
struct stu {
    string s;
    int score;
};
vector<stu> v;
int n, m;
string s;
int score;
bool cmp1(stu s1, stu s2) {
    return s1.score == s2.score ? s1.s < s2.s : s1.score>s2.score;
}
int main() {
    cin >> n >> m;

    for (int i = ; i < n; i++) {
        cin >> s >> score;
        getchar();
        stu s1;
        s1.s = s;
        s1.score = score;
        v.push_back(s1);
    }
    for (int i = ; i <= m; i++) {
        vector<stu> ans;
        string query;
        int num;
        cin >> num >> query;
        printf("Case %d: %d %s\n", i, num, query.c_str());
        int flag = ;
        if (num == ) {
            for (int j = ; j < n; j++) {
                if (v[j].s[] == query[]) {
                    ans.push_back(v[j]);
                    flag = ;
                }
            }
        }
        else if (num == ) {
            int total = , count = ;
            for (int j = ; j < n; j++) {
                if (v[j].s.substr(, ) == query) {
                    total += v[j].score;
                    count++;
                    flag = ;
                }
            }
            if(flag==)printf("%d %d\n", count, total);
        }
        else if (num == ) {
            unordered_map<string, int> mp3;
            for (int j = ; j < n; j++) {
                if (v[j].s.substr(, ) == query) {
                    mp3[v[j].s.substr(, )]++;
                    flag = ;
                }
            }
            if (flag == ) {
                for (auto it:mp3) {
                    ans.push_back({ it.first,it.second });
                }
            }
        }
        sort(ans.begin(), ans.end(), cmp1);
        for (int j = ; j < ans.size(); j++) {
            printf("%s %d\n", ans[j].s.c_str(), ans[j].score);
        }
        if (flag == )printf("NA\n");
    }
    system("pause");
}

注意点：测试3要用 unordered_map ，才能保证不超时，map会超时。

第二个小技巧，结构体和比较函数可以多用，都是数值和字符串的比较。

第三个小技巧是结果存到一个新数组里，再排序可能可以节省一点时间

第四点，输出用 printf 能节省时间，尽量不用cout