#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#include<iostream>
#include<map>
#include<string>
using namespace std;
#define rep(i,a,n) for(int i=a;i<=n;++i)
#define per(i,a,n) for(int i=n;i>=a;--i)
typedef long long ll;
typedef unsigned long long ull;
const int inf=1<<29,N=10010,M=300010;
int head[N],ver[N],edge[N],Next[N],d[N];
int n,m,k,s,t,tot,maxflow;
queue<int>q;
void add(int x,int y,int z) {
ver[++tot]=y;
edge[tot]=z;
Next[tot]=head[x];
head[x]=tot;
ver[++tot]=x;
edge[tot]=0;
Next[tot]=head[y];
head[y]=tot;
}
bool bfs() { //在残量网络上构造分层图
memset(d,0,sizeof(d));
while(q.size())q.pop();
q.push(s);
d[s]=1;
while(q.size()) {
int x=q.front();
q.pop();
for(int i=head[x]; i; i=Next[i])
if(edge[i]&&!d[ver[i]]) {
q.push(ver[i]);
d[ver[i]]=d[x]+1;
if(ver[i]==t)return 1;
}
}
return 0;
}
int dinic(int x,int flow) { //在当前分层图上增广
if(x==t)return flow;
int rest=flow,k;
for(int i=head[x]; i; i=Next[i])
if(edge[i]&&d[ver[i]]==d[x]+1) {
k=dinic(ver[i],min(rest,edge[i]));
if(!k)d[ver[i]]=0;
edge[i]-=k;
edge[i^1]+=k;
rest-=k;
}
return flow-rest;
}
string str1,str2;
map<string,int>mp;
int cnt;
int main() {
while(scanf("%d", &n)!=EOF) {
cnt=1;
mp.clear();
tot=1;
memset(head,0,sizeof(head));
s=0,t=1;
rep(i,1,n){
cin>>str1;
if(!mp[str1])mp[str1]=++cnt;
add(s,mp[str1],1);
}
scanf("%d", &m);
rep(i,1,m){
cin>>str1>>str2;
// mp[str1]=++cnt;
if(!mp[str2]){
mp[str2]=++cnt;
}
add(mp[str2],t,1);
//add(mp[str1],t,1);
//add(mp[str2],mp[str1],1);
}
scanf("%d", &k);
rep(i,1,k){
cin>>str1>>str2;
if(!mp[str1])mp[str1]=++cnt;
if(!mp[str2])mp[str2]=++cnt;
//这里必须是inf而不是1，因为某个插头可能可以由其他许多插头转接而来，这样以来，流量就不仅仅为1了
add(mp[str2],mp[str1],inf);
}
maxflow=0;
int flow=0;
while(bfs())
while(flow=dinic(s,inf))maxflow+=flow;
cout<<m-maxflow<<endl;
}
return 0;
}
撰写