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 (≤104) 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
Thinking
Simulation problem,
is to investigate the use of sort, and the map
corresponding to the three cases a student information output level corresponding to
the sum of the second output corresponds to the student site achievement
number corresponding to the site and the site corresponding to the output date 3 students
and is intended to sort title Yaoan output
this question should pay attention to the time complexity of the design code, test point 4 changed an hour ,, attention to unordered_map (but repeat their mistakes point to do some useless operation)
the specific reason has been pointed out timeouts in time-out code.
Correct codes
#include<bits/stdc++.h>
using namespace std;
struct Node {
string id;
int score;
string site, data;
Node(string id, int score, string site, string data): id(id), score(score), site(site), data(data) {
}
};
struct query {
string site;
int num;
};
vector<Node> v;
bool cmpA(const Node &a, const Node &b) {
if(a.score != b.score) return a.score > b.score;
else return a.id < b.id;
}
bool cmpC(const query &a, const query &b) {
if(a.num != b.num) return a.num > b.num;
else return a.site < b.site;
}
int main()
{
int n, m;
scanf("%d%d", &n, &m);
string id;
int score;
for(int i = 0; i < n; i++) {
cin >> id >> score;
//Node node = Node(id, score, id.substr(0, 1), id.substr(1, 3), id.substr(4, 6));
v.push_back({id, score, id.substr(1, 3), id.substr(4, 6)});
}
int type;
string term;
int cnt = 1;
while(m--) {
scanf("%d", &type);
cin >> term;
vector<Node> temp;
int sum = 0;
unordered_map<string, int> ma;
int len = term.size();
for(int i = 0; i < n; i++) {
if(len == 1 && v[i].id[0] == term[0]) temp.push_back(v[i]);
else if(len == 3 && v[i].site == term) {
sum += v[i].score;
temp.push_back(v[i]);
}
else if(len == 6 && v[i].data == term) {
ma[v[i].site]++;
}
}
printf("Case %d: %d %s\n", cnt++, type, term.c_str());
if(temp.size() == 0 && type != 3 || ma.size() == 0 && type == 3) {
printf("NA\n");
continue;
}
if(type == 1) {
sort(temp.begin(), temp.end(), cmpA);
for(int i = 0; i < temp.size(); i++) {
printf("%s %d\n", temp[i].id.c_str(), temp[i].score);
}
} else if(type == 2) {
printf("%d %d\n", temp.size(), sum);
} else if(type == 3) {
//cout << ma.size() << endl;
vector<query> q;
for(auto it : ma) { //
q.push_back({it.first, it.second});
}
sort(q.begin(), q.end(), cmpC);
for(int i = 0; i < q.size(); i++) {
printf("%s %d\n", q[i].site.c_str(), q[i].num);
}
}
}
return 0;
}
Error Code (timeout)
#include<bits/stdc++.h>
using namespace std;
struct Node {
string id;
int score;
};
struct query {
string site;
int num;
};
vector<Node> v;
bool cmpA(Node a, Node b) {
if(a.score != b.score) return a.score > b.score;
else return a.id < b.id;
}
bool cmpC(query a, query b) {
if(a.num != b.num) return a.num > b.num;
else return a.site < b.site;
}
int main()
{
int n, m;
scanf("%d%d", &n, &m);
string id;
int score;
for(int i = 0; i < n; i++) {
cin >> id >> score;
v.push_back({id, score});
}
int type;
string term;
int cnt = 1;
while(m--) {
scanf("%d", &type);
cin >> term;
vector<Node> temp;
int len = term.size();
for(int i = 0; i < n; i++) { // 此处是错误的位置
if(len == 1 && v[i].id[0] == term[0]) temp.push_back(v[i]);
else if(len == 3 && v[i].id.substr(1, 3) == term) temp.push_back(v[i]); // 此处就可以算出总和
else if(len == 6 && v[i].id.substr(4, 6) == term) temp.push_back(v[i]); //此处就需要将所求值加入到 unordered_map 不应该再重复计算
}
printf("Case %d: %d %s\n", cnt++, type, term.c_str());
if(temp.size() == 0) {
printf("NA\n");
continue;
}
if(type == 1) {
sort(temp.begin(), temp.end(), cmpA);
for(int i = 0; i < temp.size(); i++) {
printf("%s %d\n", temp[i].id.c_str(), temp[i].score);
}
} else if(type == 2) {
int sum = 0;
for(int i = 0; i < temp.size(); i++) {
sum += temp[i].score;
}
printf("%d %d\n", temp.size(), sum);
} else if(type == 3) {
vector<query> q;
map<string, int> ma;
for(int i = 0; i < temp.size(); i++) {
ma[temp[i].id.substr(1, 3)]++;
}
for(auto it = ma.begin(); it != ma.end(); it++) { //
q.push_back({it->first, it->second});
}
sort(q.begin(), q.end(), cmpC);
for(int i = 0; i < q.size(); i++) {
printf("%s %d\n", q[i].site.c_str(), q[i].num);
}
}
}
return 0;
}