POJ1087(网络流，二分图匹配）

• 题意：一间房间里面有多重插头，每种类型的插头只有一个，每种设备对应一个插头，部分设备对应的插头没有，但给出了一些不限制数量的插头转换器，求出不能工作的设备的最小数量。

• 思路：ac的第一道二分图匹配的题。二分图匹配就是特殊的网络流问题，所以用网络流的模板就可以轻松解决。虽然属于入门的网络流，因为构图问题仍然wa掉了两次，还需多练习，代码依然写的太渣。

• 代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <cmath>
#include <vector>
#include <queue>
#include <algorithm>
#include <map>
#include <set>


using namespace std;


const int INF = 0x3f3f3f3f;
const int MAX_V = 300;

struct edge {
    int to, cap, rev;
};

vector<edge> G[MAX_V];

bool used[MAX_V];

void add_edge(int from, int to, int cap) {
    G[from].push_back((edge){to, cap, G[to].size()});
    G[to].push_back((edge){from, 0, G[from].size() - 1});
}

int dfs(int v, int t, int f) {
    if(v == t) return f;
    used[v] = true;
    for(int i = 0; i < G[v].size(); i++) {
        edge &e = G[v][i];
        if(!used[e.to] && e.cap > 0) {
            int d = dfs(e.to, t, min(f, e.cap));
            if(d > 0) {
                e.cap -= d;
                G[e.to][e.rev].cap += d;
                return d;
            }
        }
    }
    return 0;
}

int max_flow(int s, int t) {
    int flow = 0;
    for(;;) {
        memset(used, 0, sizeof(used));
        int f = dfs(s, t, INF);
        if(f > 0) flow += f;
        else {
            return flow;
        }
    }
}

set<string> rec;
set<string> dev;
map<string, string> dev_rec;
vector<pair<string, string> > adapters;
map<string, int> rec2id;
map<string, int> dev2id;
map<string, bool> is_exit;

//const int MAX_N = 100 + 5;

int N, M, K;

int main() {
    while(cin >> N) {
        //initial
        for(int i = 0; i < MAX_V; i++) {
            G[i].clear();
        }
        rec.clear();
        dev.clear();
        dev_rec.clear();
        adapters.clear();
        rec2id.clear();
        dev2id.clear();
        is_exit.clear();
        //input

        for(int i = 0; i < N; i++) {
            string receptancle;
            cin >> receptancle;
            rec.insert(receptancle);
            is_exit[receptancle] = true;
        }

        cin >> M;
        for(int i = 0; i < M; i++) {
            string device, receptancle;
            cin >> device >> receptancle;
            dev.insert(device);
            rec.insert(receptancle);
            dev_rec[device] = receptancle;
        }

        cin >> K;
        for(int i = 0; i < K; i++) {
            string rec_in, rec_out;
            cin >> rec_in >> rec_out;
            rec.insert(rec_in);
            rec.insert(rec_out);
            adapters.push_back(make_pair(rec_in, rec_out));
        }

        set<string>::iterator it;
        int i = 0;
        int n = rec.size(), m = dev.size();
        for(it = rec.begin(), i = 0; it != rec.end(); it++, i++) {
            rec2id[*it] = i;
        }

        for(it = dev.begin(), i = 0; it != dev.end(); it++, i++) {
            dev2id[*it] = n + i;
        }

        int s = n + m , t = n + m + 1;

        //make graph
        for(it = rec.begin(); it != rec.end(); it++) {
            if(is_exit[*it])
                add_edge(rec2id[*it], t, 1);
        }

        for(it = dev.begin(); it != dev.end(); it++) {
            add_edge(s, dev2id[*it], 1);
        }

        for(int i = 0; i < adapters.size(); i++) {
            add_edge(rec2id[adapters[i].first], rec2id[adapters[i].second], INF);
        }

        map<string, string>::iterator d_it;
        for(d_it = dev_rec.begin(); d_it != dev_rec.end(); d_it++) {
            add_edge(dev2id[d_it->first], rec2id[d_it->second], 1);
        }

        /*DEBUG cout << "debug:" << n << " " << m << endl; for(int i = 0; i <= t; i++) { for(int j = 0; j < G[i].size(); j++) { cout<< "debug:" << i << " " << G[i][j].to << " " << G[i][j].cap << endl; } } DEBUG*/
        int flow = max_flow(s, t);
        int ans = M - flow;
        cout << ans << endl;
    }
    return 0;
}