题干：

You are in charge of setting up the press room for the inaugural meeting of the United Nations Internet eXecutive (UNIX), which has an international mandate to make the free flow of information and ideas on the Internet as cumbersome and bureaucratic as possible. 
Since the room was designed to accommodate reporters and journalists from around the world, it is equipped with electrical receptacles to suit the different shapes of plugs and voltages used by appliances in all of the countries that existed when the room was built. Unfortunately, the room was built many years ago when reporters used very few electric and electronic devices and is equipped with only one receptacle of each type. These days, like everyone else, reporters require many such devices to do their jobs: laptops, cell phones, tape recorders, pagers, coffee pots, microwave ovens, blow dryers, curling 
irons, tooth brushes, etc. Naturally, many of these devices can operate on batteries, but since the meeting is likely to be long and tedious, you want to be able to plug in as many as you can. 
Before the meeting begins, you gather up all the devices that the reporters would like to use, and attempt to set them up. You notice that some of the devices use plugs for which there is no receptacle. You wonder if these devices are from countries that didn't exist when the room was built. For some receptacles, there are several devices that use the corresponding plug. For other receptacles, there are no devices that use the corresponding plug. 
In order to try to solve the problem you visit a nearby parts supply store. The store sells adapters that allow one type of plug to be used in a different type of outlet. Moreover, adapters are allowed to be plugged into other adapters. The store does not have adapters for all possible combinations of plugs and receptacles, but there is essentially an unlimited supply of the ones they do have.

Input

The input will consist of one case. The first line contains a single positive integer n (1 <= n <= 100) indicating the number of receptacles in the room. The next n lines list the receptacle types found in the room. Each receptacle type consists of a string of at most 24 alphanumeric characters. The next line contains a single positive integer m (1 <= m <= 100) indicating the number of devices you would like to plug in. Each of the next m lines lists the name of a device followed by the type of plug it uses (which is identical to the type of receptacle it requires). A device name is a string of at most 24 alphanumeric 
characters. No two devices will have exactly the same name. The plug type is separated from the device name by a space. The next line contains a single positive integer k (1 <= k <= 100) indicating the number of different varieties of adapters that are available. Each of the next k lines describes a variety of adapter, giving the type of receptacle provided by the adapter, followed by a space, followed by the type of plug.

Output

A line containing a single non-negative integer indicating the smallest number of devices that cannot be plugged in.

Sample Input

4 
A 
B 
C 
D 
5 
laptop B 
phone C 
pager B 
clock B 
comb X 
3 
B X 
X A 
X D 

Sample Output

1

题目大意：

在一个会议室里有n种插座，每种插座一个,每个插座只能插一种以及一个电器（或者适配器）,有m个电器，每个电器有一个插头需要插在相应一种插座上,不是所有电器都能在会议室找到相应插座,有k种适配器，每种适配器可以有无限多数量,每种适配器(a, b)可以把b类插座变为a类插座,问最后有多少个电器无法使用。

解题报告：

好像直接建图并不难、、、读清楚题就好了。

AC代码：

#include<cstring>
#include<cstdio>
#include<algorithm>
#include<iostream>
#include<map>
using namespace std;
int tot;
const int INF = 0x3f3f3f3f;
struct Edge {int to,ne,w;
} e[100005 * 2];
int head[10005];
int st,ed;
int dis[10050],q[10005];//一共多少个点跑bfs，dis数组和q数组就开多大。 
void add(int u,int v,int w) {e[++tot].to=v;e[tot].w=w;e[tot].ne=head[u];head[u]=tot;
}
bool bfs(int st,int ed) {memset(dis,-1,sizeof(dis));int front=0,tail=0;q[tail++]=st;dis[st]=0;while(front<tail) {int cur = q[front];if(cur == ed) return 1;front++;for(int i = head[cur]; i!=-1; i = e[i].ne) {if(e[i].w&&dis[e[i].to]<0) {q[tail++]=e[i].to;dis[e[i].to]=dis[cur]+1;}}}if(dis[ed]==-1) return 0;return 1;
}
int dfs(int cur,int limit) {//limit为源点到这个点的路径上的最小边权 if(limit==0||cur==ed) return limit;int w,flow=0;for(int i = head[cur]; i!=-1; i = e[i].ne) {		if(e[i].w&&dis[e[i].to]==dis[cur]+1) {w=dfs(e[i].to,min(limit,e[i].w));e[i].w-=w;e[i^1].w+=w;flow+=w;limit-=w;if(limit==0) break;}}if(!flow) dis[cur]=-1;return flow;
}
int dinic() {int ans = 0;while(bfs(st,ed)) ans+=dfs(st,0x7fffffff);return ans;
}
map<string,int> mp,di;
int id,n,m,k;
int main() 
{char s1[55],s2[55];tot=1;memset(head,-1,sizeof head);cin>>n;for(int i = 1; i<=n; i++) {scanf("%s",s1);if(mp.find(s1) == mp.end()) mp[s1] = ++id;}int chatou = id;st = 0;cin>>m;for(int i = 1; i<=m; i++) {scanf("%s%s",s1,s2);if(di.find(s1) == di.end()) di[s1] = ++id;if(mp.find(s2) == mp.end()) mp[s2] = ++id;add(di[s1],mp[s2],1);add(mp[s2],di[s1],0);add(st,di[s1],1);add(di[s1],st,0);}cin>>k;for(int i = 1; i<=k; i++) {scanf("%s%s",s1,s2);add(mp[s1],mp[s2],INF);add(mp[s2],mp[s1],0);}ed = id+1;for(int i = 1; i<=chatou; i++) add(i,ed,1),add(ed,i,0);printf("%d\n",m - dinic());return 0;
}