1153 Decode Registration Card of PAT (25 分)

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

排序题
自己写的超时了一组数据，但是算了一下，极限数据应该也不会超时啊，
然后各种改，优化。。。。。

#include <bits/stdc++.h>

using namespace std;
int n,m,f;
string st;
struct Node
{
    string s, s1,s2,s3, c;
    int sorce;
    friend bool operator < (const Node &a, const Node &b){
        if(a.sorce == b.sorce)
            return a.s < b.s;
        return a.sorce > b.sorce;
    }
}node[10005];

struct EDG
{
    string ss;
    int shu;
    friend bool operator < (const EDG &a, const EDG &b){
        if(a.shu == b.shu){
            return a.ss < b.ss;
        }
        return a.shu > b.shu;
    }
};

int main(){
    cin >> n >> m;
    for(int i = 0 ; i < n ; ++ i){
        cin >> node[i].s >> node[i].sorce;
        node[i].c += node[i].s[0];
        node[i].s1 = node[i].s.substr(1,3);
        node[i].s2= node[i].s.substr(4,6);
        node[i].s3 = node[i].s.substr(10,3);
    }
    for(int i = 0; i < m; ++ i){
        cin >> f >> st;
        bool flag = true;
        printf("Case %d: %d %s
", i+1, f, st.c_str());
        vector<EDG> edg;
        if(f == 1){
            for(int j = 0; j < n; ++j){
                if(node[j].c == st){
                    flag = false;
                    edg.push_back({node[j].s,node[j].sorce});
                }
            }
        }else if(f == 2){
            int ans = 0, all = 0;
            for(int j = 0; j < n; ++j){
                if(node[j].s1 == st){
                    ans++;
                    all += node[j].sorce;
                }
            }
            if(ans){
                flag = false;
                printf("%d %d
", ans, all);
            }
        }else if(f==3){
            map<string, int> mp;
            for(int j = 0; j < n; ++j){
                if(node[j].s2 == st){
                    mp[node[j].s1] ++;
                }
            }
            for(auto it: mp) edg.push_back({it.first, it.second});
        }
        sort(edg.begin(), edg.end());
        for(int j = 0 ; j < edg.size(); ++j){
            printf("%s %d
", edg[j].ss.c_str(), edg[j].shu);
            flag = false;
        }
        if(flag)
            puts("NA");
    }
    return 0;
}