ZOJ3601-Unrequited Love

Unrequited Love

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Time Limit: 16 Seconds      Memory Limit: 131072 KB

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There are n single boys and m single girls. Each of them may love none, one or several of other people unrequitedly and one-sidedly. For the coming q days, each night some of them will come together to hold a single party. In the party, if someone loves all the others, but is not loved by anyone, then he/she is called king/queen of unrequited love.

Input

There are multiple test cases. The first line of the input is an integer T ≈ 50 indicating the number of test cases.

Each test case starts with three positive integers no more than 30000 -- n m q. Then each of the next n lines describes a boy, and each of the next m lines describes a girl. Each line consists of the name, the number of unrequitedly loved people, and the list of these people's names. Each of the last q lines describes a single party. It consists of the number of people who attend this party and their names. All people have different names whose lengths are no more than 20. But there are no restrictions that all of them are heterosexuals.

Output

For each query, print the number of kings/queens of unrequited love, followed by their names in lexicographical order, separated by a space. Print an empty line after each test case. See sample for more details.

Sample Input

2
2 1 4
BoyA 1 GirlC
BoyB 1 GirlC
GirlC 1 BoyA
2 BoyA BoyB
2 BoyA GirlC
2 BoyB GirlC
3 BoyA BoyB GirlC
2 2 2
H 2 O S
He 0
O 1 H
S 1 H
3 H O S
4 H He O S

Sample Output

0
0
1 BoyB
0

0
0

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Author: WU, Zejun
Contest: The 9th Zhejiang Provincial Collegiate Programming Contest

题意：给出一些人名和各自喜欢的人，然后有q个咨询，每个咨询给你一个聚会名单，如果其中有人喜欢着所有其他的人，且不被任何其他人喜欢，那么就符合条件，按字典序输出符合条件的所有人名。

解题思路：模拟，如有这样的人，那么必然有且只有一个

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

using namespace std;

const int INF=0x3f3f3f3f;
#define LL long long
const int MAX=30090;

map<string,int>mp;
set<int>s[MAX];
int sum;

int n,m,q;
string x[MAX];

int get(string ch)
{
    if(!mp.count(ch))
        mp[ch]=sum++;
    return mp[ch];
}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d %d %d",&n,&m,&q);
        sum=0;
        mp.clear();
        for(int i=0;i<MAX;i++)
            s[i].clear();
        string ch;
        int a,k;
        for(int i=0;i<n+m;i++)
        {
            cin>>ch>>a;
            k=get(ch);
            for(int j=0;j<a;j++)
            {
                cin>>ch;
                s[k].insert(get(ch));
            }
        }
        while(q--)
        {
            scanf("%d",&a);
            for(int i=0;i<a;i++)
                cin>>x[i];
            string p=x[0];
            int k=0;
            for(int i=1;i<a;i++)
            {
                if(!s[get(x[k])].count(get(x[i]))||s[get(x[i])].count(get(x[k])))
                    p=x[i],k=i;
            }
            for(int i=0;i<k;i++)
            {
                if(!s[get(x[k])].count(get(x[i]))||s[get(x[i])].count(get(x[k])))
                {
                    p="";
                    break;
                }
            }
            if(p=="") printf("0\n");
            else cout<<1<<" "<<p<<endl;
        }
        printf("\n");
    }
    return 0;
}