。。暴力枚举答案跑最大流。。
可以二分枚举的。。

//author: CHC
//First Edit Time:	2014-08-11 13:55
//Last Edit Time:	2014-08-11 15:39
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <set>
#include <vector>
#include <map>
#include <queue>
#include <set>
#include <algorithm>
#include <limits.h>
using namespace std;
typedef long long LL;
const int MAXN=1e+3;
const int MAXM=1e+5;
const int INF = INT_MAX;
struct Edge
{
    int from,to,ci,next;
    Edge(){}
    Edge(int _from,int _to,int _ci,int _next):from(_from),to(_to),ci(_ci),next(_next){}
}e[MAXM];
int head[MAXN],tot;
int dis[MAXN];
int top,sta[MAXN],cur[MAXN];
inline void init(){
    memset(head,-1,sizeof(head));
    tot=0;
}
inline void AddEdge(int u,int v,int ci0,int ci1=0){
    e[tot]=Edge(u,v,ci0,head[u]);
    head[u]=tot++;
    e[tot]=Edge(v,u,ci1,head[v]);
    head[v]=tot++;
}
inline bool bfs(int st,int et){
    memset(dis,0,sizeof(dis));
    dis[st]=1;
    queue <int> q;
    q.push(st);
    while(!q.empty()){
        int now=q.front();
        q.pop();
        for(int i=head[now];i!=-1;i=e[i].next){
            int next=e[i].to;
            if(e[i].ci&&!dis[next]){
                dis[next]=dis[now]+1;
                if(next==et)return true;
                q.push(next);
            }
        }
    }
    return false;
}
LL Dinic(int st,int et){
    LL ans=0;
    while(bfs(st,et)){
        //printf("here\n");
        top=0;
        memcpy(cur,head,sizeof(head));
        int u=st,i;
        while(1){
            if(u==et){
                int pos,minn=INF;
                //printf("top:%d\n",top);
                for(i=0;i<top;i++)
                {
                    if(minn>e[sta[i]].ci){
                        minn=e[sta[i]].ci;
                        pos=i;
                    }
                    //printf("%d --> %d\n",e[sta[i]].from,e[sta[i]].to);
                }
                for(i=0;i<top;i++){
                    e[sta[i]].ci-=minn;
                    e[sta[i]^1].ci+=minn;
                }
                top=pos;
                u=e[sta[top]].from;
                ans+=minn;
                //printf("minn:%d\n\n",minn);
            }
            for(i=cur[u];i!=-1;cur[u]=i=e[i].next)
                if(e[i].ci&&dis[u]+1==dis[e[i].to])break;
            if(cur[u]!=-1){
                sta[top++]=cur[u];
                u=e[cur[u]].to;
            }
            else {
                if(top==0)break;
                dis[u]=0;
                u=e[sta[--top]].from;
            }
        }
    }
    return ans;
}
int path[MAXN],cnt[MAXN];
int Find(int x){
    return x==path[x]?x:path[x]=Find(path[x]);
}
bool Union(int x,int y){
    x=Find(x);y=Find(y);
    if(x==y)return true;
    path[y]=x;
    cnt[x]+=cnt[y];
    cnt[y]=0;
    return true;
}
set <int> ss[200];
set <int> ::iterator it;
int jihe[MAXN],prehead[MAXN];
int main()
{
    int t;
    int n,m,f;
    scanf("%d",&t);
    while(t--){
        scanf("%d%d%d",&n,&m,&f);
        for(int i=0;i<MAXN;i++)
            path[i]=i,cnt[i]=1;
        init();
        for(int i=0;i<200;i++)
            ss[i].clear();
        //1~100     girl
        //101~200   boy
        for(int i=0,x,y;i<m;i++){
            scanf("%d%d",&x,&y);
            ss[x].insert(y+100);
        }
        for(int i=0,x,y;i<f;i++){
            scanf("%d%d",&x,&y);
            Union(x,y);
        }
        jihe[0]=0;
        for(int i=1;i<=n;i++)
            if(path[i]==i){
                jihe[++jihe[0]]=i;
            }
            else ss[Find(i)].insert(ss[i].begin(),ss[i].end());
        int ans=1,flag=0;
        for(;ans<=100;ans++){
            //0     -->源点
            //201   -->汇点
            init();
            for(int i=1;i<=jihe[0];i++){
                for(it=ss[jihe[i]].begin();it!=ss[jihe[i]].end();it++){
                    AddEdge(jihe[i],*it,cnt[jihe[i]]);
                    //printf("%d-->%d val:%d\n",jihe[i],*it,cnt[jihe[i]]);
                }
            }
            for(int i=1;i<=jihe[0];i++)
            {
                AddEdge(0,jihe[i],ans*cnt[jihe[i]]);
                //printf("%d-->%d val:%d\n",0,jihe[i],ans*cnt[jihe[i]]);
            }
            for(int i=101;i<=100+n;i++)
            {
                AddEdge(i,201,ans);
                //printf("%d-->%d val:%d\n",i,201,ans);
            }
            //printf("ans:%d\n",ans);
            LL tt=Dinic(0,201);
            //printf("%I64d %I64d\n",tt,(LL)ans*n);
            if(tt==(LL)ans*n){
                flag=ans;
            }
            else break;
        }
        printf("%d\n",flag);
    }
    return 0;
}