1153 Decode Registration Card of PAT

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 
using namespace std;
struct node{
    char cardnums[16];
    int score;
};
struct vnode{
    int sitenum, cnts;
    vnode(int sn, int ct): sitenum(sn), cnts(ct){}
};
vector<node> vl[26];
pair<int, int> sitecnts[1024];
unordered_map<int, map<int, int>> mp;
bool cmp1(const node &a, const node &b){
    return a.score == b.score ? strcmp(a.cardnums, b.cardnums) < 0 : a.score > b.score;
}
bool cmp2(const vnode &a, const vnode &b){
    return a.cnts == b.cnts ? a.sitenum < b.sitenum : a.cnts > b.cnts;
}
int convert(char str[], int s, int e){
    int res = 0;
    while(s <= e){
        res = res * 10 + str[s++] - '0';
    }
    return res;
}
int main()
{
    int n, m;
    scanf("%d %d", &n, &m);
    for(int i = 0; i < n; ++i){
        node x;
        scanf("%s %d", x.cardnums, &x.score);
        vl[x.cardnums[0]-'A'].push_back(x);
        int sitenum = convert(x.cardnums, 1, 3);
        sitecnts[sitenum].first++;
        sitecnts[sitenum].second += x.score;
        int date = convert(x.cardnums, 4, 9);
        mp[date][sitenum]++;
    }
    sort(vl['A' - 'A'].begin(), vl['A' - 'A'].end(), cmp1);
    sort(vl['T' - 'A'].begin(), vl['T' - 'A'].end(), cmp1);
    sort(vl['B' - 'A'].begin(), vl['B' - 'A'].end(), cmp1);
    for(int i = 1; i <= m; ++i){
        int type;
        scanf("%d" ,&type);
        printf("Case %d: %d ", i, type);
        if(type == 1){
            char lev;
            scanf(" %c", &lev);
            printf("%c\n", lev);
            if(vl[lev - 'A'].size() == 0) printf("NA\n");
            else{
                for(auto &x : vl[lev - 'A']){
                    printf("%s %d\n", x.cardnums, x.score);
                }
            }
        }else if(type == 2){
            int sitenum;
            scanf("%d", &sitenum);
            printf("%d\n", sitenum);
            if(sitecnts[sitenum].first == 0) printf("NA\n");
            else printf("%d %d\n", sitecnts[sitenum].first, sitecnts[sitenum].second);
        }else{
            int date;
            scanf("%d", &date);
            printf("%06d\n", date);
            if(mp.count(date) == 0) printf("NA\n");
            else{
                vector<vnode> v;
                for(auto &x : mp[date]){
                    v.push_back(vnode(x.first, x.second));
                }
                sort(v.begin(), v.end(), cmp2);
                for(auto &x : v){
                    printf("%d %d\n", x.sitenum, x.cnts);
                }
            }
        }
    }
    return 0;
}