题目
先判断是否有解，然后开两个栈，分别存进入递归时间复杂度为O（1）的变量，和时间复杂度为O（n）的变量，最后取最大值。

#include <bits/stdc++.h>
using namespace std;
int T;
int tong[1000100];
stack <int> s, s2;
int main()
{
//  freopen("Time.txt", "w", stdout);
    scanf("%d", &T);
    while (T--)
    {
        memset(tong, 0, sizeof(tong)); 
        while (s.size()) s.pop();
        int L, flag = 0, F1 = 0, E1 = 0, ans = 0, now = 0; string TI;//flag为2则不进入 
        cin >> L >> TI;  int len = TI.size();
        if (TI[2] == '1')
            ans = 0;
        else 
            for (int i = 4; i < len; i++)
            {
                if (TI[i] >= '0' && TI[i] <= '9')
                    ans = ans * 10 + TI[i] - '0';
            }
        for (int i = 1; i <= L; i++)
        {       
            char c1;
            cin >> c1;
            if (c1 == 'F')
            {   
                F1++; 
                string st, ed; char bianliang;
                cin >> bianliang >> st >> ed;
                if (tong[bianliang - '0'])//如果已经进入了该变量，就判断ERR 
                    flag = 1;//以后的都不计入时间复杂度 (
                if (st[0] == 'n' && ed[0] != 'n' && flag != 1)
                {
                    /*if (ed[0] == 'n')
                        now += 0;
                    else if (!flag)
                        now++;*/
                    flag = 2;
                    tong[bianliang - '0'] = 0;
                }   
                else if (st[0] != 'n') //说明此时st是数 
                {
                    int sta = 0, eda = 0;
                    for (int i = 0; i < st.size(); i++)
                        if (st[i] <= '9' && st[i] >= '0')
                            sta = sta * 10 + st[i] - '0';
                    if (ed[0] == 'n' && !flag)
                    {
                        tong[bianliang - '0'] = 1;
                        s.push(bianliang - '0');    
                        int SI = s.size();
                        now = max(now, SI);
                    }
                    else
                    {
                        for (int i = 0; i < ed.size(); i++)
                            if (ed[i] <= '9' && ed[i] >= '0')
                                eda = eda * 10 + ed[i] - '0';
                        if (sta > eda && flag != 1) flag = 2, s.push(214748);
                        if (sta <= eda) tong[bianliang - '0'] = 1, s2.push(bianliang - '0');// s2表示进去但不加时间复杂度的 
                    }
                }
            }
            else
            {
                E1++; int ha = 0; 
                if (s.size() )
                    ha = s.top(), s.pop();
                else if (!s.size() && s2.size() ) 
                    ha = s2.top(), s2.pop();
                tong[ha] = 0;
                if (ha == 214748 && flag != 1)
                    flag = 0;
                    
            }
        }
        if (F1 != E1 || flag == 1)
            printf("ERR\n");
        else 
         {
            if (ans != now)
                printf("No\n", ans, now);
            else 
            printf("Yes\n");    
         }
    }
    return 0;
}