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

Then

After the info of testees, there are

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。。。。。

 1 #include <bits/stdc++.h>
 2 
 3 using namespace std;
 4 int n,m,f;
 5 string st;
 6 struct Node
 7 {
 8     string s, s1,s2,s3, c;
 9     int sorce;
10     friend bool operator < (const Node &a, const Node &b){
11         if(a.sorce == b.sorce)
12             return a.s < b.s;
13         return a.sorce > b.sorce;
14     }
15 }node[10005];
16 
17 struct EDG
18 {
19     string ss;
20     int shu;
21     friend bool operator < (const EDG &a, const EDG &b){
22         if(a.shu == b.shu){
23             return a.ss < b.ss;
24         }
25         return a.shu > b.shu;
26     }
27 };
28 
29 int main(){
30     cin >> n >> m;
31     for(int i = 0 ; i < n ; ++ i){
32         cin >> node[i].s >> node[i].sorce;
33         node[i].c += node[i].s[0];
34         node[i].s1 = node[i].s.substr(1,3);
35         node[i].s2= node[i].s.substr(4,6);
36         node[i].s3 = node[i].s.substr(10,3);
37     }
38     for(int i = 0; i < m; ++ i){
39         cin >> f >> st;
40         bool flag = true;
41         printf("Case %d: %d %s
", i+1, f, st.c_str());
42         vector<EDG> edg;
43         if(f == 1){
44             for(int j = 0; j < n; ++j){
45                 if(node[j].c == st){
46                     flag = false;
47                     edg.push_back({node[j].s,node[j].sorce});
48                 }
49             }
50         }else if(f == 2){
51             int ans = 0, all = 0;
52             for(int j = 0; j < n; ++j){
53                 if(node[j].s1 == st){
54                     ans++;
55                     all += node[j].sorce;
56                 }
57             }
58             if(ans){
59                 flag = false;
60                 printf("%d %d
", ans, all);
61             }
62         }else if(f==3){
63             map<string, int> mp;
64             for(int j = 0; j < n; ++j){
65                 if(node[j].s2 == st){
66                     mp[node[j].s1] ++;
67                 }
68             }
69             for(auto it: mp) edg.push_back({it.first, it.second});
70         }
71         sort(edg.begin(), edg.end());
72         for(int j = 0 ; j < edg.size(); ++j){
73             printf("%s %d
", edg[j].ss.c_str(), edg[j].shu);
74             flag = false;
75         }
76         if(flag)
77             puts("NA");
78     }
79     return 0;
80 }