网络流，设备、插头和转接器建图(简单map的应用)

题意：

给你n个插座，m个设备，每台设备都有对应的插座，有k个转接器。

要求：求满足不能插上插座的用电器最少个数 

solution:

HINT：每种适配器都有无限个，所以建图的时候要改为INF。

答案为m-idnic()

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<vector>
#include<queue>
#include<map>
#include<cstring>
#define mp make_pair
#define pb push_back
#define first fi
#define second se
#define pw(x) (1ll << (x))
#define sz(x) ((int)(x).size())
#define all(x) (x).begin(),(x).end()
#define rep(i,l,r) for(int i=(l);i<(r);i++)
#define per(i,r,l) for(int i=(r);i>=(l);i--)
#define FOR(i,l,r) for(int i=(l);i<=(r);i++)
#define eps 1e-9
#define PIE acos(-1)
#define cl(a,b) memset(a,b,sizeof(a))
#define fastio ios::sync_with_stdio(false);cin.tie(0);
#define lson l , mid , ls
#define rson mid + 1 , r , rs
#define ls (rt<<1)
#define rs (ls|1)
#define INF 0x3f3f3f3f
#define LINF 0x3f3f3f3f3f3f3f3f
#define freopen freopen("in.txt","r",stdin);
#define cfin ifstream cin("in.txt");
#define lowbit(x) (x&(-x))
#define sqr(a) a*a
#define ll long long
#define ull unsigned long long
#define vi vector<int>
#define pii pair<int, int>
#define dd(x) cout << #x << " = " << (x) << ", "
#define de(x) cout << #x << " = " << (x) << "
"
#define endl "
"
using namespace std;
int n,m,k;
const int maxn=505;
map<string,int>ids; 

struct Edge{
    int u,v,cap;
};
int S,T;
vi G[maxn];
int dep[maxn]; 
vector<Edge> edge;
inline int id(string s)
{
    int m=sz(ids);
    if(ids.count(s))return ids[s];
    return ids[s]=m; 
}

void addedge(int u,int v,int cap)
{
    edge.pb((Edge){u,v,cap});
    edge.pb((Edge){v,u,0});
    int t=sz(edge);
    G[u].pb(t-2);
    G[v].pb(t-1);
}

bool bfs()
{
    cl(dep,-1);
    int Q[maxn]; 
    int h=0,t=1;dep[S]=0;Q[t]=S;
    while(h<t){
//        de(T);
        int x=Q[++h];
        if(x==T)return 1;
        rep(i,0,sz(G[x])){
            Edge& e=edge[G[x][i]];
//            de(e.v);puts("here");
            if(dep[e.v]==-1&&e.cap){
                dep[e.v]=dep[x]+1;
                Q[++t]=e.v;
            }
        }
    }
    return 0;
}
int dfs(int x,int f)
{
    if(x==T)return f;
    int used=0,t;
    rep(i,0,sz(G[x])){
        Edge& e=edge[G[x][i]];
//        dd(e.u),de(e.v);
        if(e.cap&&dep[e.v]==dep[x]+1){
            t=dfs(e.v,min(e.cap,f));
            e.cap-=t;edge[G[x][i]^1].cap+=t;
            used+=t;f-=t;
            if(!f)return used;
        }
    }
    if(!used)dep[x]=-1;
    return used;
}
int dinic()
{
    int ans=0;
    while(bfs())ans+=dfs(S,INF);
    return ans;
}
void mapping()
{
    S=0;
    string in[maxn],out[maxn];id("S");
    cin>>n;
    FOR(i,1,n)cin>>in[i];
    cin>>m;
    FOR(i,n+1,n+m)cin>>in[i]>>out[i],addedge(0,id(in[i]),1),addedge(id(in[i]),id(out[i]),1);
    cin>>k;
    FOR(i,n+m+1,n+m+k)cin>>in[i]>>out[i],addedge(id(in[i]),id(out[i]),INF);
//    FOR(i,1,n)dd(in[i]),de(id(in[i]));
    T=n+m+k+1;
    FOR(i,1,n)addedge(id(in[i]),T,1);
}
int main()
{
    mapping();
    cout<<m-dinic()<<endl;
    return 0;
}