ZOJ 3207 80ers' Memory(水题）

80ers' Memory

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Time Limit: 1 Second      Memory Limit: 32768 KB

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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

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

 

 

map秒过了。。。

水题。。

#include<stdio.h>

#include<iostream>

#include<string.h>

#include<string>

#include<algorithm>

#include<map>

using namespace std;

map<string,int>mp;

int main()

{

    int n,k;

    int m;

    string str;

    while(scanf("%d",&n)!=EOF)

    {

        mp.clear();

        while(n--)

        {

            cin>>str;

            if(mp[str]==0)mp[str]=1;

        }

        int ans;

        scanf("%d",&k);

        while(k--)

        {

            scanf("%d",&m);

            ans=0;

            while(m--)

            {

                cin>>str;

                ans+=mp[str];

            }

            printf("%d\n",ans);

        }

    }

    return 0;

}