POJ-1087 A Plug for UNIX

题目链接：POJ-1087 A Plug for UNIX

题意

有$n$个插座，m个用电器，每个用电器都对应一个插头(插头类型与插座相同则可插入充电)，$k$种转换器(每种数量无限)，$s_1$　$s_2$表示一个$s_2$类型插座可以转换成一个$s_1$类型插座，问最少有几个用电器无法同时充电。

---

思路

可建图用最大流求解，源点指向$n$个插座，$m$个用电器指向汇点，插座指向对应的用电器，容量均为1，对于一个转换器$(s_1,s_2)$，插座$s_2$指向$s_1$，容量为无穷大。最大流表示可同时充电的最大充电器数量，用$m$减去最大流即为答案。

---

代码实现

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <queue>
#include <map>
using std::queue;
using std::map;
using std::cin;
using std::cout;
using std::endl;
using std::string;
const int INF = 0x3f3f3f3f, N = 500, M = 1000;
int head[N], d[N];
int s, t, tot, maxflow;
struct Edge
{
    int to, cap, nex;
} edge[M];
queue<int> q;
void add(int x, int y, int z) {
    edge[++tot].to = y, edge[tot].cap = z, edge[tot].nex = head[x], head[x] = tot;
    edge[++tot].to = x, edge[tot].cap = 0, edge[tot].nex = head[y], head[y] = tot;
}
bool bfs() {
    memset(d, 0, sizeof(d));
    while (q.size()) q.pop();
    q.push(s); d[s] = 1;
    while (q.size()) {
        int x = q.front(); q.pop();
        for (int i = head[x]; i; i = edge[i].nex) {
            int v = edge[i].to;
            if (edge[i].cap && !d[v]) {
                q.push(v);
                d[v] = d[x] + 1;
                if (v == t) return true;
            }
        }
    }
    return false;
}
int dinic(int x, int flow) {
    if (x == t) return flow;
    int rest = flow, k;
    for (int i = head[x]; i && rest; i = edge[i].nex) {
        int v = edge[i].to;
        if (edge[i].cap && d[v] == d[x] + 1) {
            k = dinic(v, std::min(rest, edge[i].cap));
            if (!k) d[v] = 0;
            edge[i].cap -= k;
            edge[i^1].cap += k;
            rest -= k;
        }
    }
    return flow - rest;
}
void init() {
    tot = 1, maxflow = 0;
    s = 0, t = 1;
    memset(head, 0, sizeof(head));
}

int main() {
    int n, m, k;
    while (cin >> n) {
        init();
        int idx = 1;
        string s1, s2;
        map<string, int> mp;
        for (int i = 0; i < n; i++) {
            cin >> s1;
            if (!mp[s1]) mp[s1] = ++idx;
            add(0, mp[s1], 1);
        }
        cin >> m;
        for (int i = 0; i < m; i++) {
            cin >> s1 >> s2;
            if (!mp[s1]) mp[s1] = ++idx;
            if (!mp[s2]) mp[s2] = ++idx;
            add(mp[s2], mp[s1], 1);
            add(mp[s1], 1, 1);
        }
        cin >> k;
        for (int i = 0; i < k; i++) {
            cin >> s1 >> s2;
            if (!mp[s1]) mp[s1] = ++idx;
            if (!mp[s2]) mp[s2] = ++idx;
            add(mp[s2], mp[s1], INF);
        }
        while (bfs()) maxflow += dinic(s, INF);
        cout << (m - maxflow) << endl;
    }
    return 0;
}