zoj 4704 Unrequited Love 贪心 孤独者

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  

链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=4704

题意：n+m个人，各自喜欢num个人。  然后q个询问，每个询问给你num个人，问你在这num个人中，是否有个孤独者喜欢这num个人中的所有人，但是这num个人都没喜欢他。

做法：首先明确的是，这里只可能有一个人符合要求，否者 条件矛盾 。先假设这个人 是0号，从从小到大枚举 这num个人，假设第i个人是这个孤独者，那么i+j个人如果喜欢i，或者i不喜欢i+j，那么i就不符合条件，接下来就假设这个孤独者是i+j。i+j之前可以断定不是孤独者。 到最后就可以得到假设的孤独者，然后从0开始再循环一遍，判断他是否真是孤独者。然后就ok了。

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <malloc.h>
#include <ctype.h>
#include <math.h>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;
#include <stack>
#include <queue>
#include <vector>
#include <deque>
#include <set>
#include <map>
using namespace std;
#define maxn 505

map<int,set<int> > my;
map<string,int> turn;

int ID(string& str)
{
	if(turn.count(str)==0)
		return turn[str]=turn.size();
	return turn[str];

}
string str[40000];
int main()
{
	int t;
	int i,j;
	int n,m,q;
	int num;
	cin>>t;
	while(t--)
	{
		my.clear();
		turn.clear();
		cin>>n>>m>>q;
		for(i=0;i<n+m;i++)
		{
			string a,b;;
			int n;
			int numa,numb;
			cin>>a;
			numa=ID(a);
			cin>>num;
			for(j=0;j<num;j++)
			{
				cin>>b;
				numb=ID(b);
				my[numa].insert(numb);
			}
		}

		for(i=0;i<q;i++)
		{
			string a,b,nw;
			cin>>num;
			int id=1;
			for(j=0;j<num;j++)
			{
				cin>>str[j];
				if(j==0)
				{
					nw=str[j];
					continue;
				}
				else
				{
					int num_nw=turn[nw];
					int num_a=turn[str[j]];
					if(my[num_nw].count(num_a)==0||my[num_a].count(num_nw))
					{
						nw=str[j];
					}
				}
			}

			int num_nw=turn[nw];
			for(j=0;j<num;j++)
			{
				if(str[j]==nw)
					continue;
				int num_a=turn[str[j]];
				if(my[num_nw].count(num_a)==0||my[num_a].count(num_nw))
				{
					nw="";
					break;
				}
			}
			if(nw=="")
				cout<<0<<endl;
			else
				cout<<1<<' '<<nw<<endl;
		}
		cout<<endl;
	}

	return 0;
}