NOIP2017 time complexity of large simulation

Write a large simulation questions.

Since the limit is written, the code is equivalent to the examination room, a group of chaos.

Topic link: P3952 time complexity

---

Credited first few lessons:

• Compare the size of the string of numbers honestly write function, lexicographical have made a mistake a few times
• Stack empty when not only pop () will RE, top () will access the top of the stack RE
• Digital read as a character into consideration more than one digit

---

Ideas:

1. I use the online approach.

2. The memory structure using a loop, including a variable name, start and end, whether the n-related, is able to enter (this trouble, whether pushed on the stack with n-related and whether to enter the line)

3. Use of a variable stores the current into the layers (containing only referred to n), when pressed into a + "and n-related", when a pop-up "and the related n" -, takes a maximum value

4. Use a bool whether the current cycle into the memory. Cancel "can not enter the" pop-up when a state can not enter the structure (if the "inaccessible" state, the structure will not be referred to as "inaccessible")

Source:

//MiserWeyte is now "mzWyt"
#include <bits/stdc++.h>
using namespace std;
int n, tgt, rel, curr = 0, maxx;
bool used[30];
string tar;
bool err, notin;
struct sts{
    char tpe, nam;
    string sta, end; 
    bool use, nin;
};
bool le(string a, string b){
    if(a=="n" || b=="n" ) return false;
    int numa = atoi(a.c_str());
    int numb = atoi(b.c_str());
    return numa > numb;
}
stack <sts> s;
void init(){
    err = false;
    memset(used, 0, sizeof(used));
    curr = 0;
    maxx = 0;
    notin = false;
    while(s.size()) s.pop();
}
int main(){
    int t;
    cin >> t;
    while(t--){
        init();
        cin >> n >> tar;
        tgt = 0;
        if(tar[2] != '1'){
            for(int i=0; i<tar.length(); i++){
                if(tar[i] >= '0' && tar[i] <= '9'){
                    tgt *= 10;
                    tgt += tar[i] - '0';
                }
            }
        }
        for(int i=0; i<n; i++){
            sts temp;
            cin >> temp.tpe;
            if(temp.tpe == 'F'){
                cin >> temp.nam ;
                cin>> temp.sta; 
                cin>> temp.end;
                temp.use = false;
                temp.nin = false;
                if(used[temp.nam - 'a']){
                    err = true;
//                  break;
                }
                used[temp.nam - 'a'] = true;
                if(notin){
                    s.push(temp);
                }
                else if(temp.sta == "n" && temp.end == "n") s.push(temp);
                else if(temp.sta != "n" && temp.end == "n"){
                    temp.use = true;
                    curr ++;
                    maxx = max(maxx, curr);
                    s.push(temp);
                }
                else if((temp.sta == "n" && temp.end != "n") || le(temp.sta, temp.end)){
                    notin = true;
                    temp.nin = true;
                    s.push(temp);
                }
                else s.push(temp);
            }
            else{
                if(s.empty()){
                    err = true;
                    continue;
                } 
                if(s.size() && s.top().use && !s.top().nin) curr --;
//              cout << s.top().use;
                if(s.size() && s.top().nin) notin = false;
                used[s.top().nam - 'a'] = false;
                if(s.size()) s.pop();
                
            }
//          cout << "curr" << curr << endl;
//          if(notin) cout << "NOTIN" << endl;
        }
        if(s.size()) err = true;
        if(err) cout << "ERR\n";
        else{
            if(maxx == tgt) cout << "Yes\n";
            else cout << "No\n";
//          cout << curr << endl;
        }
    }
}