ZOJ 3826

Hierarchical Notation

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Time Limit: 2 Seconds      Memory Limit: 131072 KB

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In Marjar University, students in College of Computer Science will learn EON (Edward Object Notation), which is a hierarchical data format that uses human-readable text to transmit data objects consisting of attribute-value pairs. The EON was invented by Edward, the headmaster of Marjar University.

The EON format is a list of key-value pairs separated by comma ",", enclosed by a couple of braces "{" and "}". Each key-value pair has the form of "<key>":"<value>". <key> is a string consists of alphabets and digits. <value> can be either a string with the same format of <key>, or a nested EON.

To retrieve the data from an EON text, we can search it by using a key. Of course, the key can be in a nested form because the value may be still an EON. In this case, we will use dot "." to separate different hierarchies of the key.

For example, here is an EON text:

{"headmaster":"Edward","students":{"student01":"Alice","student02":"Bob"}}

 

• For the key "headmaster", the value is "Edward".
• For the key "students", the value is {"student01":"Alice","student02":"Bob"}.
• For the key "students"."student01", the value is "Alice".

 

As a student in Marjar University, you are doing your homework now. Please write a program to parse a line of EON and respond to several queries on the EON.

Input

There are multiple test cases. The first line of input contains an integer T indicating the number of test cases. For each test case:

The first line contains an EON text. The number of colons ":" in the string will not exceed 10000 and the length of each key and non-EON value will not exceed 20.

The next line contains an integer Q (0 <= Q <= 1000) indicating the number of queries. Then followed by Q lines, each line is a key for query. The querying keys are in correct format, but some of them may not exist in the EON text.

The length of each hierarchy of the querying keys will not exceed 20, while the total length of each querying key is not specified. It is guaranteed that the total size of input data will not exceed 10 MB.

Output

For each test case, output Q lines of values corresponding to the queries. If a key does not exist in the EON text, output "Error!" instead (without quotes).

Sample Input

1
{"hm":"Edward","stu":{"stu01":"Alice","stu02":"Bob"}}
4
"hm"
"stu"
"stu"."stu01"
"students"

Sample Output

"Edward"
{"stu01":"Alice","stu02":"Bob"}
"Alice"
Error!

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Author: LU, Yi
Source: The 2014 ACM-ICPC Asia Mudanjiang Regional Contest

赛后看见别人补这个题，我便也补，其实场上没有看这个题，听说是到模拟，我觉得正解应该就是字符串处理来模拟数据库，然而这个题可以更简单，由于数据少，10000条key，所以hush便出现重的几率很小，看见大家都是字符串+hush水的，我也这么做的，总之，这样做简单啊。。。

注意边界处理

#include <cstdio>
#include <iostream>
#include <sstream>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <algorithm>
using namespace std;
#define ll long long
#define _cle(m, a) memset(m, a, sizeof(m))
#define repu(i, a, b) for(int i = a; i < b; i++)
#define repd(i, a, b) for(int i = b; i >= a; i--)
#define sfi(n) scanf("%d", &n)
#define sfl(n) scanf("%lld", &n)
#define pfi(n) printf("%d\n", n)
#define pfl(n) printf("%lld\n", n)
#define MAXN 1000005
#define BASE 131
char s[MAXN];
map<ll, pair<int, int> > ma;
ll id(char c)
{
    return ((ll)c) % BASE;
}
int p;
void Create(ll f)
{
    while(s[p] != '}')
    {
        if(s[++p] == '}') return ;
        ll t = f;
        while(s[p] != ':')
        {
            t = t * BASE + id(s[p]);
            p++;
        }
        int st = ++p;
        if(s[p] == '{')
        {
            Create(t * BASE + id('.'));
        }
        else
        {
            while(s[p] != ',' && s[p] != '}') p++;
        }
        if(s[p] == '}')
        ma[t] = pair<int, int>(st, p + 1);
        else ma[t] = pair<int, int>(st, p);
        //cout<<t<<endl;
    }
    return ;
}

int main()
{
    int T;
    sfi(T);
    while(T--)
    {
        ma.clear();
        scanf("%s", s);
        int n;
        char t[MAXN];
        p = 0;
        Create(0);
        sfi(n);
        repu(i, 0, n)
        {
           scanf("%s", t);
           ll key = 0;
           int len = strlen(t);
           repu(j, 0, len) key = key * BASE + id(t[j]);
           if(!ma.count(key))
           {
               printf("Error!");
           }
           else
           {
               pair<int, int> tt = ma[key];
               repu(j, tt.first, tt.second + 1) printf("%c", s[j]);
           }
           puts("");
        }
    }
    return 0;
}