Presumably, you all have known the question of stable marriage match. A girl will choose a boy; it is similar as the game of playing house we used to play when we are kids. What a happy time as so many friends playing together. And it is normal that a fight or a quarrel breaks out, but we will still play together after that, because we are kids.
Now, there are $2n$ kids, $n$ boys numbered from $1$ to $n$, and $n$ girls numbered from $1$ to $n$. you know, ladies first. So, every girl can choose a boy first, with whom she has not quarreled, to make up a family. Besides, the girl $X$ can also choose boy $Z$ to be her boyfriend when her friend, girl $Y$ has not quarreled with him. Furthermore, the friendship is mutual, which means $a$ and $c$ are friends provided that $a$ and $b$ are friends and $b$ and $c$ are friend.
Once every girl finds their boyfriends they will start a new round of this game—marriage match. At the end of each round, every girl will start to find a new boyfriend, who she has not chosen before. So the game goes on and on.
Now, here is the question for you, how many rounds can these $2n$ kids totally play this game?
There are several test cases. First is a integer $T$, means the number of test cases.
Each test case starts with three integer $n, m$ and $f$ in a line $(3≤n≤100,0<m<n*n,0≤f<n)$. $n$ means there are $2 \times n$ children, $n$ girls(number from $1$ to $n$) and $n$ boys(number from $1$ to $n$).
Then $m$ lines follow. Each line contains two numbers $a$ and $b$, means girl $a$ and boy $b$ had never quarreled with each other.
Then $f$ lines follow. Each line contains two numbers $c$ and $d$, means girl $c$ and girl $d$ are good friends.
For each case, output a number in one line. The maximal number of Marriage Match the children can play.
- 女孩编号：$0 \sim n-1$；
- 男孩编号：$n \sim 2n-1$；
- 若上图跑出的最大流为$mid \times n$，则说明当前答案可行，
- 本题还可以用二分图匹配来做，即首先建立女孩与男孩的二分图,并且连好可能的边. 然后进行一轮匹配,如果此时有完美匹配,那么就删除这些匹配边,进行第二轮匹配,如果还有完美匹配，就继续……