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) << "\n"
 #define endl "\n"
 using namespace std;
 int n,m,k;
 const int maxn=;
 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,});
     int t=sz(edge);
     G[u].pb(t-);
     G[v].pb(t-);
 }

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