UVa 753 A Plug for UNIX (最大流)

题意：给定 n 种插座，m种设备，和k个转换器，问你最少有几台设备不能匹配。

析：一个很裸的网络流，直接上模板就行，建立一个源点s和汇点t，源点和每个设备连一条边，每个插座和汇点连一条边，然后再连转换器，

最后跑一次最大流即可。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-5;
const int maxn = 500 + 10;
const int mod = 1e6;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
  return r >= 0 && r < n && c >= 0 && c < m;
}
struct Edge{
  int from, to, cap, flow;
};

struct Dinic{
  int n, m, s, t;
  vector<Edge> edges;
  vector<int> G[maxn];
  bool vis[maxn];
  int d[maxn];
  int cur[maxn];

  void init(){
    edges.clear();
    for(int i = 0; i < maxn; ++i)  G[i].clear();
  }
  bool bfs(){
    memset(vis, 0, sizeof vis);
    queue<int> q;
    q.push(s);
    d[s] = 0;  vis[s] = true;
    while(!q.empty()){
      int x = q.front();  q.pop();
      for(int i = 0; i < G[x].size(); ++i){
        Edge &e = edges[G[x][i]];
        if(!vis[e.to] && e.cap > e.flow){
          vis[e.to] = 1;
          d[e.to] = d[x] + 1;
          q.push(e.to);
        }
      }
    }
    return vis[t];
  }

  int dfs(int x, int a){
    if(x == t || a == 0)  return a;
    int flow = 0, f;
    for(int& i = cur[x]; i < G[x].size(); ++i){
      Edge& e = edges[G[x][i]];
      if(d[x] + 1 == d[e.to] && (f = dfs(e.to, min(a, e.cap-e.flow))) > 0){
        e.flow += f;
        edges[G[x][i]^1].flow -= f;
        flow += f;
        a -= f;
        if(!a)  break;
      }
    }
    return flow;
  }

  int maxflow(int s, int t){
    this->s = s;  this->t = t;
    int flow = 0;
    while(bfs()){
      memset(cur, 0, sizeof cur);
      flow += dfs(s, INF);
    }
    return flow;
  }

  void addEdge(int from, int to, int cap){
    edges.push_back(Edge{from, to, cap, 0});
    edges.push_back(Edge{to, from, 0, 0});
    m = edges.size();
    G[from].push_back(m-2);
    G[to].push_back(m-1);
  }
};
map<string, int> mp;

int getId(const string &s){
  if(mp.count(s))  return mp[s];
  return mp[s] = mp.size();
}
vector<int> socket;
vector<P> device;
vector<P> convert;

void init(){
  mp.clear();
  socket.clear();
  device.clear();
  convert.clear();
}

int solve(){
  int s = mp.size() + 1;
  int t = mp.size() + 2;
  Dinic dinic;
  dinic.init();
  for(int i = 0; i < device.size(); ++i){
    dinic.addEdge(s, device[i].first, 1);
    dinic.addEdge(device[i].first, device[i].second, 1);
  }
  for(int i = 0; i < socket.size(); ++i)
    dinic.addEdge(socket[i], t, 1);
  for(int i = 0; i < convert.size(); ++i)
    dinic.addEdge(convert[i].first, convert[i].second, INF);
  return (int)device.size() -dinic.maxflow(s, t);
}

int main(){
  int T;  cin >> T;
  while(T--){
    cin >> n;
    init();
    string s;
    for(int i = 0; i < n; ++i){
      cin >> s;
      socket.push_back(getId(s));
    }
    cin >> n;
    string ss;
    for(int i = 0; i < n; ++i){
      cin >> s >> ss;
      device.push_back(P(getId(s), getId(ss)));
    }
    cin >> n;
    for(int i = 0; i < n; ++i){
      cin >> s >> ss;
      convert.push_back(P(getId(s), getId(ss)));
    }
    cout << solve() << endl;
    if(T) cout << endl;
  }
  return 0;
}