2014牡丹江——Hierarchical Notation
字符串模拟
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; }