2014 Mudanjiang -- hierarchical notation

Question Link

String Simulation

const int MAXN = 2000000;char ipt[MAXN], t[MAXN];int f[MAXN], len, to[MAXN];map<string, string> mp[MAXN];string x, key, ans;string i2s(int n){    string ret = "";    if (n == 0)        ret += '0';    else    {        while (n)        {            ret += n % 10 + '0';            n /= 10;        }    }    ret += '*';    reverse(ret.begin(), ret.end());    return ret;}int s2i(string& s){    int ret = 0;    FF(i, 1, s.length())        ret = ret * 10 + s[i] - '0';    return ret;}void fun_f(){    stack<int> sk;    sk.push(0);    int cnt = 0;    REP(i, len)    {        if (ipt[i] == '{')        {            sk.push(++cnt);            f[i] = cnt;            to[f[i]] = i;        }        else if (ipt[i] == '}')        {            sk.pop();            f[i] = sk.top();        }    }}void fun_ipt(){    int state = 0, lv = 0, nxt;    REP(i, len)    {        switch (ipt[i])        {        case '\"':            if (state == 0 || state == 2)                x = "\"";            else if (state == 1)                key = x + '\"';            else                mp[lv][key] = x + '\"';            state = (state + 1) % 4;            break;        case '{':            nxt = f[i];            if (state == 2)            {                mp[lv][key] = i2s(nxt);                state = 0;            }            lv = nxt;            break;        case '}':            lv = f[i];            break;        case ':' || ',':            break;        default:            x += ipt[i];        }    }}void fun_case(){    int state = 0, lv = 0, n;    RI(n);    REP(kase, n)    {        bool ok = true;        state = 0, lv = 1;        RS(t);        len = strlen(t);        REP(i, len)        {            if (t[i] == '\"')            {                if (state == 0)                    x = '\"';                else                {                    x += '\"';                    if (mp[lv].count(x) == 0)                    {                        ok = false;                        break;                    }                }                ans = mp[lv][x];                state = !state;            }            else if (t[i] == '.')            {                if (ans[0] != '*')                {                    ok = false;                    break;                }                lv = s2i(ans);            }            else                x += t[i];        }        if (ok)        {            if (ans[0] == '*')            {                int i = to[s2i(ans)], cnt = 0;                while (true)                {                    if (ipt[i] == '{')                        cnt++;                    else if (ipt[i] == '}')                    {                        if (--cnt == 0)                            break;                    }                    putchar(ipt[i]);                    i++;                }                puts("}");            }            else                puts(ans.c_str());        }        else            puts("Error!");    }}int main(){    int T;    RI(T);    while (T--)    {        REP(i, MAXN)            mp[i].clear();        RS(ipt);        len = strlen(ipt);        fun_f();        fun_ipt();        fun_case();    }    return 0;}

2014 Mudanjiang -- hierarchical notation