在C ++中显示餐厅的食物订单表

假设我们有一个数组订单，它代表顾客在餐厅里完成的订单。因此，orders [i] = [cust_namei，table_numi，food_itemi]，其中cust_namei是客户名称，table_numi是客户表号，而food_itemi是商品客户订单。

我们必须退还餐厅的“展示桌”。这里的“显示表”是一个表，其行条目表示每个表订购的每种食品有多少。第一列为表格编号，其余各列按字母顺序对应于每个食品。第一行应为标题，其第一列为“表格”，然后是食品名称。

因此，如果输入像orders = [["Amal","3","Paratha"],["Bimal","10","Biryni"],["Amal","3","Fried Chicken"],["Raktim","5","Water"],["Raktim","5","Paratha"],["Deepak","3","Paratha"]]，则输出将是[["Table","Biryni","Fried Chicken","Paratha","Water"],["3","0","1","2","0"],["5","0","0","1","1"],["10","1","0","0","0"]]]]

例 

让我们看下面的实现以更好地理解-

#include <bits/stdc++.h>
using namespace std;
void print_vector(vector<vector<string> > v){
   cout << "[";
   for(int i = 0; i<v.size(); i++){
      cout << "[";
      for(int j = 0; j <v[i].size(); j++){
         cout << v[i][j] << ", ";
      }
      cout << "],";
   }
   cout << "]"<<endl;
}
typedef long long int lli;
class Solution {
public:
   vector <string> split(string& s, char delimiter){
      vector <string> tokens;
      string token;
      istringstream tokenStream(s);
      while(getline(tokenStream, token, delimiter)){
         tokens.push_back(token);
      }
      return tokens;
   }
   static bool cmp(vector <string>& a, vector <string>& b){
      lli an = stol(a[0]);
      lli bn = stol(b[0]);
      return an < bn;
   }
   vector<vector<string>> displayTable(vector<vector<string>>& o) {
      map <string, map < string, int> >m;
      set <string> names;
      set <string> t;
      for(auto &it : o){
         t.insert(it[1]);
         bool ok = true;
         if(!ok){
            vector <string> v = split(it[2], ' ');
            for(auto& x : v){
               m[it[1]][x]++;
               names.insert(x);
            }
         }
         else{
            m[it[1]][it[2]]++;
            names.insert(it[2]);
         }
      }
      vector < vector <string> > ret;
      vector <string> temp(names.begin(), names.end());
      temp.insert(temp.begin(), "Table");
      ret.push_back(temp);
      for(auto& it : t){
         vector <string> te;
         te.push_back(it);
         for(auto& x : names){
            te.push_back(to_string(m[it][x]));
         }
         ret.push_back(te);
      }
      sort(ret.begin() + 1, ret.end(), cmp);
      return ret;
   }
};
main(){
   Solution ob;
   vector<vector<string>> v = {{"Amal","3","Paratha"},{"Bimal","10","Biryni"},{"Amal","3","Fried
Chicken"},{"Raktim","5","Water"},{"Raktim","5","Paratha"},{"Deepak"," 3","Paratha"}};
   print_vector(ob.displayTable(v));
}

输入值

{{"Amal","3","Paratha"},{"Bimal","10","Biryni"},{"Amal","3","Fried Chicken"},{"Raktim","5","Water"},{"Raktim","5","Paratha"},{"Deepak"," 3","Paratha"}}

输出结果

[[Table, Biryni, Fried Chicken, Paratha, Water, ],[3, 0, 1, 2, 0, ],[5, 0, 0, 1, 1, ],[10, 1, 0, 0, 0, ],]