Marriage Match II

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1608 Accepted Submission(s): 566

Problem Description

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?

Input

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*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.

Output

For each case, output a number in one line. The maximal number of Marriage Match the children can play.

Sample Input

       1 4 5 2 1 1 2 3 3 2 4 2 4 4 1 4 2 3 

Sample Output

2

题意：n个女生和n个男生，女生和男生没吵过架就能在一起。。

女生和女生之间仅仅要是好朋友，那么没吵过架的男生就都能在一起，女所有选择好了要在一起的男生就算一次匹配，然后再来。可是女生不能选先前选择过的男生，问最多能玩出几次这种匹配

题解：用匈牙利算法匹配，然后把匹配到的女生和男生的关系变为吵架（这样下次她就不会选择到他了）

注意：女生A和B是朋友。B和C是朋友。那么和A、B、C没吵过架的人都能被A、B、C选择

#include 
     
      #include 
      
       using namespace std;int n,map[111][111],vis[111],d[111],v[111],fa[111];int Find(int x)	//并查集——查找祖先{
          return fa[x]==x?x:fa[x]=Find(fa[x]);}void Union(int x,int y)	//并查集——合并两点{
          x=Find(x);    y=Find(y);    if(x!=y)fa[x]=y;}int find(int x)	//匈牙利算法{
          int i;    for(i=1;i<=n;i++)    if(map[x][i])    {
              if(!vis[i])        {
                  vis[i]=1;            if(d[i]==-1||find(d[i]))            {
                      d[i]=x;                return 1;            }        }    }    return 0;}int main (void){
          int t,m,f,i,j,k,l,s;    cin>>t;    while(t--&&cin>>n>>m>>f)    {
              memset(map,0,sizeof(map));        for(i=1;i<=n;i++)        fa[i]=i;        for(i=0;i
       
        >j>>k;            map[j][k]=1;        }        for(i=0;i
        
         >k>>l;            Union(k,l);	//先用并查集把女生中是好朋友的弄到一堆        }        for(i=1;i<=n;i++)        for(j=1;j<=n;j++)        if(Find(i)==Find(j)	//假设i和j是好朋友        for(k=1;k<=n;k++)        if(map[i][k]||map[j][k])	//假设k没和i吵过架或者没和j吵过架        map[i][k]=map[j][k]=1;	//那么k能被i与j选择        s=0;        while(1)        {
                  memset(d,-1,sizeof(d));            for(i=1,k=0;i<=n;i++)            {
                      memset(vis,0,sizeof(vis));                k+=find(i);	//匈牙利算法记录匹配数            }            if(k==n)s++;	//所有都匹配了就记录为一次            else break;	//不然就结束            for(i=1;i<=n;i++)	//然后把匹配男女的关系处理为吵架            map[d[i]][i]=0;        }        cout<
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