zoj 3601 Unrequited Love(STL，灵活题)

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本文介绍了一种算法解决单恋国王问题，通过输入单身男女及其单恋对象的信息，在每场单身派对中找出单恋国王，并详细展示了算法实现过程。

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
http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=4704
#include<iostream>
#include<algorithm>
#include<string>
#include<map>
#include<set>
#include<cmath>
#include<string.h>
#include<stdlib.h>
#include<cstdio>
#define ll long long
using namespace std;
int main(){
	int t;
	cin>>t;
	while(t--){
		set<string> x[30001];  //set的 .count用法，学习 
		map<string,int> id; //把每个人转换成数字下标i，对应x[i]
		string z[30001];
		string h,w;
		int n,m,a,b;
		cin>>n>>m>>a;
		int p=-1;
		for(int i=0;i<n+m;++i){
			cin>>h>>b;
			if(id.count(h)==0)
				id[h]=++p;
			for(int j=0;j<b;++j){
				cin>>w;
				x[id[h]].insert(w);}}
		for(int i=0;i<a;++i){
			cin>>b;
			for(int j=0;j<b;++j)
				cin>>z[j];
			string g=z[0];
			int flag=0;
			for(int j=1;j<b;++j){
				if(!x[id[z[flag]]].count(z[j])||x[id[z[j]]].count(z[flag])){//如果没有进入这个if，那就说明z[flag]肯定满足，反过来即z[j]肯定不满足 
					g=z[j];  
					flag=j;}}  //如果z[flag]不满足了，那就把flag换成j，再继续看z[flag']满不满足 
			for(int j=0;j<flag;++j){
				if(!x[id[z[flag]]].count(z[j])||x[id[z[j]]].count(z[flag])){ //这一步是看z[flag]和它之前的值符不符合条件（因为上面都是往后符合的） 
					g="0";
					break;}}
			if(g=="0")
				cout<<0<<endl;
			else
				cout<<1<<" "<<g<<endl;}
		cout<<endl;}
	return 0;
}