ZOJ3207-80ers' Memory

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80ers' Memory

Time Limit: 1 Second Memory Limit: 32768 KB

I guess most of us are so called 80ers, which means that we were born in the 1980's. This group of people shared a lot of common memories. For example, the Saint Seiya, the YoYo ball, the Super Mario, and so on. Do you still remember these?

Input

There will be ONLY ONE test case.

The test case begins with a positive integer N, (N < 100).
Then there will be N lines each containing a single word describing a keyword of the typical 80ers' memories. Each word consists of letters, [a-zA-Z], numbers, [0-9], and the underline, '_'. The length of each word would be no more than 20.
Then one line follows with a positive integer K, (K < 100).
The last part of the test case will be K lines. The i-th line contains the keywords given by the i-th person and starts with a positive integer N_i. Then there will be N_i words separated by white spaces. Each of these words has no more than 20 characters.
All the words are case sensitive.

Output

For each of these K people, you are asked to count the number of typical 80ers' keywords on his/her list and output it in a single line.

Sample Input

4
Saint_Seiya
YoYo_ball
Super_Mario
HuLuWa
3
2 Saint_Seiya TiaoFangZi
1 KTV
3 HuLuWa YOYO_BALL YoYo_ball

Sample Output

1
0
2

Author: GAO, Fei
Source: The 6th Zhejiang Provincial Collegiate Programming Contest

题意：第一行为一个正整数N（1<=N<100）。以下N行每行一个关键单词，由字母，数字或下划线构成。接下来一个正整数K（1<=K<100）。以下K行每行第一个数为Ni，表示该行有Ni个单词，接下来是这Ni个单词，每个单词之间用空格分隔。对于K行的每一行，输出每行含有多少个关键单词。

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
#include <set>
#include <stack>
#include <map>
#include <climits>

using namespace std;
#define LL long long

int main()
{
    char ch[109][30],s[30];
    int n,m;
    while(~scanf("%d",&n))
    {
        for(int i=0; i<n; i++)
            scanf("%s",ch[i]);
        scanf("%d",&m);
        while(m--)
        {
            int a,sum=0;
            scanf("%d",&a);
            for(int i=0;i<a;i++)
            {
                scanf("%s",s);
                for(int j=0; j<n; j++)
                    if(strcmp(s,ch[j])==0) sum++;
            }
            printf("%d\n",sum);
        }
    }
    return 0;
}