【发布时间】：2017-10-03 00:09:56
【问题描述】：
#include <algorithm>
#include <fstream>
#include <iostream>
#include <vector>
using namespace std;
const int max_applications_num = 1000;
vector<string> vector_authors;
vector<string> vector_titles;
vector<string> vector_venue;
vector<int> vector_year;
vector<string> vector_presentation;
void Tokenize(string line, vector<string> &tokens, string delimiters = "\t ") {
string token = "";
string OneCharString = " ";
for (int i = 0; i < line.size(); i++)
if (find(delimiters.begin(), delimiters.end(), line[i]) !=
delimiters.end()) // line[i] is one of the delimiter characters
{
if (token != "")
tokens.push_back(token);
token = "";
} else {
OneCharString[0] = line[i];
token += OneCharString;
}
if (token != "")
tokens.push_back(token);
}
void SaveApplication(const vector<string> &tokens) {
string authors = tokens[1];
string title = tokens[2];
string venue = tokens[3];
int year = atoi(tokens[4].c_str());
string presentation = tokens[5];
vector_authors.push_back(authors);
vector_titles.push_back(title);
vector_venue.push_back(venue);
vector_year.push_back(year);
vector_presentation.push_back(presentation);
// cout << "in save" << endl;
}
void remove_application(int pos) {
vector_authors.erase(vector_authors.begin() + pos);
vector_titles.erase(vector_titles.begin() + pos);
vector_venue.erase(vector_venue.begin() + pos);
vector_year.erase(vector_year.begin() + pos);
vector_presentation.erase(vector_presentation.begin() + pos);
// cout << "in remove" << endl;
}
void sort() {
for (int j = 0; j <= vector_year.size() - 1; j++) {
int temp1 = vector_year.at(j);
int i = j - 1;
while (i > -1 and vector_year.at(i) > temp1) {
vector_year.at(i + 1) = vector_year.at(i);
i = i - 1;
}
vector_year.at(i + 1) = temp1;
}
for (int j = 0; j <= vector_authors.size() - 1; j++) {
string temp2 = vector_authors.at(j);
int i = j - 1;
while (i > -1 and vector_authors.at(i) > temp2) {
vector_authors.at(i + 1) = vector_authors.at(i);
i = i - 1;
}
vector_authors.at(i + 1) = temp2;
}
for (int j = 0; j <= vector_titles.size() - 1; j++) {
string temp3 = vector_titles.at(j);
int i = j - 1;
while (i > -1 and vector_titles.at(i) > temp3) {
vector_titles.at(i + 1) = vector_titles.at(i);
i = i - 1;
}
vector_titles.at(i + 1) = temp3;
}
for (int j = 0; j <= vector_venue.size() - 1; j++) {
string temp4 = vector_venue.at(j);
int i = j - 1;
while (i > -1 and vector_venue.at(i) > temp4) {
vector_venue.at(i + 1) = vector_venue.at(i);
i = i - 1;
}
vector_venue.at(i + 1) = temp4;
}
for (int j = 0; j <= vector_presentation.size() - 1; j++) {
string temp5 = vector_presentation.at(j);
int i = j - 1;
while (i > -1 and vector_presentation.at(i) > temp5) {
vector_presentation.at(i + 1) = vector_presentation.at(i);
i = i - 1;
}
vector_presentation.at(i + 1) = temp5;
}
// cout << "in sort" << endl;
}
void print() {
for (int i = 0; i < vector_authors.size(); i++) {
cout << vector_authors.at(i) << "\t" << vector_titles.at(i) << "\t"
<< "\t" << vector_venue.at(i) << "\t" << vector_year.at(i) << "\t" << vector_presentation.at(i) << endl;
}
cout << "\n" << endl;
}
void ExecuteCommands(const char *fname) {
ifstream inf;
inf.open(fname);
string line;
while (getline(inf, line).good()) {
vector<string> tokens;
Tokenize(line, tokens, "\t ");
if (tokens.size() == 0)
continue;
if (tokens[0].compare("save_application") == 0)
SaveApplication(tokens);
else if (tokens[0].compare("remove_application") == 0)
remove_application(atoi(tokens[1].c_str()));
else if (tokens[0].compare("sort") == 0)
sort();
else if (tokens[0].compare("print") == 0)
print();
}
inf.close();
}
int main(int argc, char **argv) {
if (argc != 2) {
cout << "usage: executable.o command.txt\n";
return 1;
}
ExecuteCommands(argv[1]);
return 0;
}
所以，这是我在学校做的一个实验室的代码。我们应该将某些元素放入一个向量中，打印这些向量，对它们排序，再次打印它们，删除一个向量，最后一次打印它们。对于我们的排序，我们需要根据出版年份对它们进行排序。
"authors_list1" "title1" "conference1" 2016 "poster"
"authors_list3" "title3" "conference2" 2010 "oral"
"authors_list2" "title2" "journal1" 2015 "none"
所以，当我排序时，我得到了这个：
"authors_list1" "title1" "conference1" 2010 "none"
"authors_list2" "title2" "conference2" 2015 "oral"
"authors_list3" "title3" "journal1" 2016 "poster"
这是预期的输出：
"authors_list3" "title3" "conference2" 2010 "oral"
"authors_list2" "title2" "journal1" 2015 "none"
"authors_list1" "title1" "conference1" 2016 "poster"
年份的顺序是正确的，但其他一切的顺序都不正确。我需要我的其他元素随着岁月的流逝而效仿。有什么办法吗？
附：对于本实验，我们不允许使用类或结构。这是我所有的代码。
【问题讨论】：
-
与您的问题无关，但不是多个向量，您为什么不使用 结构 和单个向量来包含它？实际上可能会使排序更容易一些。
-
我们不允许。实验室没有指定结构或类。
-
根据你对遵守法律条文和精神的感觉有多接近，你可以使用元组。
-
请编辑您的问题以包含预期（以及实际）输出。或其他重要或相关的详细信息。
-
@J.Khelly Please read this answer。您的数据属于“笨拙”类别。您所需要的只是一个索引数组。不需要结构或类。