时间复杂度

#include<bits/stdc++.h>
using namespace std;
int n;
string k;
int str_to_int(string a)
{
    int ans=0;
    for(int i=0;i<a.length();i++)
        ans=ans*10+a[i]-'0';
    return ans;
}
struct oppo{
    char na;
    int s;
    oppo(char na,int s):na(na),s(s){};
};
stack<oppo> v;
bool MAP[205];
void work()
{
    int sj=0,ans=0,all=0,flag=0;
    while(v.size()) v.pop();
    memset(MAP,0,sizeof(MAP));

    cin>>n>>k;
    if(k!="O(1)"){
        for(int i=0;i<k.length();i++)
            if('0'<=k[i]&&k[i]<='9')
                sj=sj*10+k[i]-'0';
    }
    char t,a;
    string b,c;
    for(int i=1;i<=n;i++)
    {
        cin>>t;
        if(t=='F')
        {
            cin>>a>>b>>c;
            int bb,cc;
            if(b=="n") bb=2020;
            else bb=str_to_int(b);

            if(c=="n") cc=2020;
            else cc=str_to_int(c);

            if(MAP[a]) flag=1;
            MAP[a]=1;

            if(cc-bb>100){
                v.push(oppo(a,all));
                all++;
                ans=max(all,ans);
            }
            else{
                if(bb>cc){
                    v.push(oppo(a,all));
                    all=-10000;
                }else{
                    v.push(oppo(a,all));
                }
            }
        }
        else
        {
            if(v.size()){
                all=v.top().s;
                MAP[v.top().na]=0;
                v.pop();
            }else
                flag=1;
        }
    }
    if(flag||v.size()) puts("ERR");
    else if(ans==sj) puts("Yes");
    else puts("No");
    return ;
}
int T;
int main()
{
    cin>>T;
    while(T--)
        work();
    return 0;
}