A Plug for UNIX POJ

题意：

几种插头，每一种都只有一个，但有无限个插头转换器，转换器（a，b） 意味着 可以把b转换为a，有几个设备，每个设备对应一种插头，求所不能匹配插头的设备数量

这个题可以用二分图做 ， 我用的是最大流，最后用设备数 减去 最大匹配数即可

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <map>
#include <queue>
#define mem(a, b) memset(a, b, sizeof(a))
using namespace std;
const int maxn = 100100, INF = 0x7fffffff;
int  head[maxn], d[maxn], cur[maxn];
int n, m, s, t, z;
int cnt;
map<string, int> mapp;
struct node{
    int u, v, c, next;
}Node[maxn*4];

void  add_(int u, int v, int c)
{
    Node[cnt].u = u;
    Node[cnt].v = v;
    Node[cnt].c = c;
    Node[cnt].next = head[u];
    head[u] = cnt++;
}

void add(int u, int v, int c)
{
    add_(u, v, c);
    add_(v, u, 0);
}

bool bfs()
{
    queue<int> Q;
    mem(d, 0);
    Q.push(s);
    d[s] = 1;
    while(!Q.empty())
    {
        int u = Q.front(); Q.pop();
        for(int i=head[u]; i!=-1; i=Node[i].next)
        {
            node e = Node[i];
            if(!d[e.v]  && e.c > 0)
            {
                d[e.v] = d[e.u] + 1;
                Q.push(e.v);
            //    cout<< u << "   " << e.v << "   " << t <<endl;
                if(e.v == t) break;
            }
        }
    }
    return d[t] != 0;
}

int dfs(int u, int cap)
{
 //   cout<< cap <<endl;
    if(u == t || cap == 0)
        return cap;
    int ret = 0;
    for(int &i=cur[u]; i!=-1; i=Node[i].next)
    {
        node e = Node[i];
        if(d[e.v] == d[e.u] + 1 && e.c > 0)
        {
            int V = dfs(e.v, min(cap, e.c));
            Node[i].c -= V;
            Node[i^1].c += V;
            cap -= V;
            ret += V;
          //  cout<< V <<endl;
            if(cap == 0) break;
        }
    }
    return ret;
}

int dinic()
{
    int ans = 0;
    while(bfs())
    {
        memcpy(cur, head, sizeof(head));
        ans += dfs(s, INF);
     //   cout<< ans <<endl;
    }
    return ans;
}
int  main()
{
    cin>> n;
    mem(head, -1);
    s = 0, t = n + 3*m + 2*z + 10;
    for(int i=1; i<=n; i++)
    {
        string str;
        cin>> str;
        mapp[str] = i;
        add(s, i, 1);
    }
    cin>> m;
    for(int i=1; i<=m; i++)
    {
        string str, sstr;
        cin>> str >> sstr;
        add(n+i, n+m+i, 1);
        add(n+m+i, t, INF);
        if(!mapp[sstr])
        {
            mapp[sstr] = n + m*2 + i;
        }
        add(mapp[sstr], n+i, INF);
    }
    cin>> z;
    for(int i=1; i<=z; i++)
    {
        string u, v;
        cin>> u >> v;
        if(!mapp[u])
            mapp[u] = n+3*m+i;
        if(!mapp[v])
            mapp[v] = n+3*m+z+i;
  //     add(s, mapp[v], INF);
        add(mapp[v], mapp[u], INF);
    }
 //   cout<< m <<  "  " <<dinic() <<endl;
    cout<< m - dinic() <<endl;

    return 0;
}