关于 省赛模拟赛（迪迦桑专场）

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?

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.

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.

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

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

#include <stdio.h>
#include <string.h>
#define N 250

int main()
{
    int T, n, m, k, i, j, q, x;
    char a[N][N], ch[N];
    int b[N][N];
    scanf("%d", &T);
    while(T--)
    {
        memset(b,0,sizeof(b));
        scanf("%d%d",&n,&m);
        scanf("%d",&k);
        for (i=0; i<k; i++)
        {
            scanf("%s",a[i]);
        }
        for (i=0; i<m; i++)
        {
            scanf("%d",&x);
            for (j=0; j<x; j++)
            {
                scanf("%s", ch);
                for (q=0; q<k; q++)
                {
                    if (strcmp(ch, a[q])==0)
                    {
                        b[i][q]=1;
                        break;
                    }
                }
            }
        }
        int vi[N];
        int f, qq;
        for (i=0; i<n; i++)
        {
            for (j=0; j<m; j++)
            {
                scanf("%d",&vi[j]);
            }
            int sum=0;
            for (j=0; j<k; j++)///就是这里的k一不小心敲成了x导致了一直WA；
            {
                f=1;
                for (q=0; q<m; q++)
                {
                    if (vi[q]!=b[q][j])
                    {
                        f=0;
                        break;
                    }
                }
                if (f==1)
                {
                    sum++;
                    qq=j;
                }
            }
            if(sum==1)
                printf("%s\n",a[qq]);
            else
                printf("Let‘s go to the library!!\n");
        }
    }
    return 0;
}