浙江省赛--C

What Kind of Friends Are You?

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Time Limit: 1 Second      Memory Limit: 65536 KB

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Japari Park is a large zoo home to extant species, endangered species, extinct species, cryptids and some legendary creatures. Due to a mysterious substance known as Sandstar, all the animals have become anthropomorphized into girls known as Friends.

Kaban is a young girl who finds herself in Japari Park with no memory of who she was or where she came from. Shy yet resourceful, she travels through Japari Park along with Serval to find out her identity while encountering more Friends along the way, and eventually discovers that she is a human.

However, Kaban soon finds that it's also important to identify other Friends. Her friend, Serval, enlightens Kaban that she can use some questions whose expected answers are either "yes" or "no" to identitfy a kind of Friends.

To be more specific, there are n Friends need to be identified. Kaban will ask each of them q same questions and collect their answers. For each question, she also gets a full list of animals' names that will give a "yes" answer to that question (and those animals who are not in the list will give a "no" answer to that question), so it's possible to determine the name of a Friends by combining the answers and the lists together.

But the work is too heavy for Kaban. Can you help her to finish it?

Input

There are multiple test cases. The first line of the input is an integer T (1 ≤ T ≤ 100), indicating the number of test cases. Then T test cases follow.

The first line of each test case contains two integers n (1 ≤ n ≤ 100) and q (1 ≤ q ≤ 21), indicating the number of Friends need to be identified and the number of questions.

The next line contains an integer c (1 ≤ c ≤ 200) followed by c strings p₁, p₂, ... , p_c (1 ≤ |p_i| ≤ 20), indicating all known names of Friends.

For the next q lines, the i-th line contains an integer m_i (0 ≤ m_i ≤ c) followed by m_i strings s_{i, 1}, s_{i, 2}, ... , s_{i, mi} (1 ≤ |s_{i, j}| ≤ 20), indicating the number of Friends and their names, who will give a "yes" answer to the i-th question. It's guaranteed that all the names appear in the known names of Friends.

For the following n lines, the i-th line contains q integers a_{i, 1}, a_{i, 2}, ... , a_{i, q} (0 ≤ a_{i, j} ≤ 1), indicating the answer (0 means "no", and 1 means "yes") to the j-th question given by the i-th Friends need to be identified.

It's guaranteed that all the names in the input consist of only uppercase and lowercase English letters.

Output

For each test case output n lines. If Kaban can determine the name of the i-th Friends need to be identified, print the name on the i-th line. Otherwise, print "Let's go to the library!!" (without quotes) on the i-th line instead.

Sample Input

2
3 4
5 Serval Raccoon Fennec Alpaca Moose
4 Serval Raccoon Alpaca Moose
1 Serval
1 Fennec
1 Serval
1 1 0 1
0 0 0 0
1 0 0 0
5 5
11 A B C D E F G H I J K
3 A B K
4 A B D E
5 A B K D E
10 A B K D E F G H I J
4 B D E K
0 0 1 1 1
1 0 1 0 1
1 1 1 1 1
0 0 1 0 1
1 0 1 1 1

Sample Output

Serval
Let's go to the library!!
Let's go to the library!!
Let's go to the library!!
Let's go to the library!!
B
Let's go to the library!!
K

Hint

The explanation for the first sample test case is given as follows:

As Serval is the only known animal who gives a "yes" answer to the 1st, 2nd and 4th question, and gives a "no" answer to the 3rd question, we output "Serval" (without quotes) on the first line.

As no animal is known to give a "no" answer to all the questions, we output "Let's go to the library!!" (without quotes) on the second line.

Both Alpaca and Moose give a "yes" answer to the 1st question, and a "no" answer to the 2nd, 3rd and 4th question. So we can't determine the name of the third Friends need to be identified, and output "Let's go to the library!!" (without quotes) on the third line.

题意：给你两个整数n和q（验证的朋友人数和每次验证问的问题数量），然后输入c，在输入c个字符串（朋友的名字），然后输入每个验证题的人数和姓名，然后给出每个朋友对于问题的答案（1--是，0--否），如果能够确认朋友，输出朋友的名字，否则输出“Let's go to the library!!”。

思路：没有想到有二进制的方式去表示最后的结果，所以最后也没有做出来，看了学校大神的思路才明白。

如图所示：

将每个人的初始分数置为0，第 i 行输入的名字则其分数加2的（i-1）次方，得到每个名字最后的值，然后在输入的矩阵的每行进行如果值是1则加上2的（j-1）次方，然后进行判断，如果有唯一相同的分数就输出这个人的名字，否则输出“Let's go to the library!!”

#include <iostream>
#include <string>
#include <cstring>
#include <map>

using namespace std;

map<string,int> ok;

int main()
{
    ios::sync_with_stdio(false);
    string name[205],str;
    int m,t,n2[25],T,n,q,mb,df[205],num;
    cin>>T;
    n2[1]=1;
    for(int i=2;i<=22;i++)
        n2[i]=2*n2[i-1];
    while(T--)
    {
        cin>>n>>q;
        cin>>m;
        ok.clear();
        memset(df,0,sizeof(df));
        for(int i=1;i<=m;i++)
        {
            cin>>name[i];
            ok[name[i]]=i;
        }
        for(int i=1;i<=q;i++)
        {
            cin>>t;
            for(int j=1;j<=t;j++)
            {
                cin>>str;
                df[ok[str]]+=n2[i];
            }
        }
        for(int i=1;i<=n;i++)
        {
            num=mb=0;
            for(int j=1;j<=q;j++)
            {
                cin>>t;
                if(t)
                    mb+=n2[j];
            }
            for(int j=1;j<=m;j++)
                if(df[j]==mb)
                    num++;
            if(num==1)
            {
                for(int j=1;j<=m;j++)
                    if(df[j]==mb)
                        cout<<name[j]<<endl;
            }
            else
                cout<<"Let's go to the library!!"<<endl;
        }
    }
    return 0;
}

 

 

 

 

 

 

 

 

 

 

 

 

 

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