采用EK算法解网络流经典题，本题构图思路比较明确。

//Unix会议室插座转换
//网络流-EK算法
//Time:47Ms     Memory:1188K
#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<queue>
using namespace std;

#define MAX 505
#define MAXS 25
#define INF 0x3f3f3f3f

int n,m,k;
int s,t;
int res[MAX][MAX];  //残留网络
int pre[MAX];
int len;    //插头种类数
int no[MAX];    //插头下标编号
char trans[MAX][MAXS];  //插头名称

int getNum(char *str)   //插头名称->编号
{
    for(int i = 0; i < len; i++)
    {
        if(!strcmp(trans[i], str))
            return no[i];
    }
    return -1;
}

bool bfs()
{
    memset(pre,-1,sizeof(pre));
    queue<int> q;
    q.push(s);  pre[s] = 0;
    while(!q.empty()){
        int cur = q.front();
        q.pop();
        for(int i = 1; i <= t; i++)
        {
            if(pre[i] == -1 && res[cur][i])
            {
                pre[i] = cur;
                if(i == t)  return true;
                q.push(i);
            }
        }
    }
    return false;
}

int EK()
{
    int maxFlow = 0;
    while(bfs())
    {
        int mind = INF;
        for(int i = t; i != s; i = pre[i])
            mind = min(mind, res[pre[i]][i]);
        for(int i = t; i != s; i = pre[i])
        {
            res[pre[i]][i] -= mind;
            res[i][pre[i]] += mind;
        }
        maxFlow += mind;
    }
    return maxFlow;
}

int main()
{
	//freopen("in.txt", "r", stdin);

    memset(res,0,sizeof(res));
    scanf("%d", &n);
    s = 0;  t = n+1;    //汇点在不断更新
    len = 0;
    for(int i = 1; i <= n; i++)
    {
        no[len] = i;
        scanf("%s", trans[len++]);
    }
    scanf("%d", &m);
    t += m;
    int num[2];
    char str[2][MAXS];
    for(int i = n+1; i <= n+m; i++)
    {
        scanf("%s%s", str[0], str[1]);
        int num = getNum(str[1]);
        if(num == -1)
        {
            no[len] = num = t++;
            strcpy(trans[len++], str[1]);
        }
        res[s][i] = res[i][num] = 1;
    }
    scanf("%d", &k);
    for(int i = 0; i < k; i++)
    {
        scanf("%s%s", str[0], str[1]);
        for(int j = 0; j < 2; j++){
            num[j] = getNum(str[j]);
            if(num[j] == -1)
            {
                no[len] = num[j] = t++;
                strcpy(trans[len++], str[j]);
            }
        }
        res[num[0]][num[1]] = INF;
    }
    //汇点已固定-更新汇点的邻接边
    for(int i = 1; i <= n; i++)
        res[i][t] = 1;
    printf("%d\n", m - EK());

	return 0;
}