「BZOJ 1711」Dining - 网络流

每一头牛只喜欢吃一些食品和饮料而别的一概不吃。农夫做了 F 1 \leq F \leq 100 )种食品并准备了 D 1 \leq D \leq 100 )种饮料。 N 1 \leq N \leq 100 )头牛都以决定了是否愿意吃某种食物和喝某种饮料。

农夫想给每一头牛一种食品和一种饮料,使得尽可能多的牛得到喜欢的食物和饮料。

每一件食物和饮料只能由一头牛来用。

链接

BZOJ 1711

题解

将一头牛拆成两个点 A A' ,从每个食品向能吃它的牛的 A 点连一条边,从每头牛的 A' 点向它能喝的饮料连一条边。

源点向每个食品连边,每个饮料向汇点连边。

边权均为 1 ,最大流即为答案。

代码

#include <cstdio>
#include <climits>
#include <queue>
#include <algorithm>

const int MAXN = 100;
const int MAXM = 100;

struct Node;
struct Edge;

struct Node {
    Edge *e, *c;
    int l;
} N[MAXN * 2 + MAXM * 2 + 2];

struct Edge {
    Node *s, *t;
    int f, c;
    Edge *next, *r;

    Edge(Node *s, Node *t, const int c) : s(s), t(t), f(0), c(c), next(s->e) {}
};

struct Dinic {
    bool makeLevelGraph(Node *s, Node *t, const int n) {
        for (int i = 0; i < n; i++) N[i].l = 0, N[i].c = N[i].e;

        std::queue<Node *> q;
        q.push(s);
        s->l = 1;

        while (!q.empty()) {
            Node *v = q.front();
            q.pop();

            for (Edge *e = v->e; e; e = e->next) if (!e->t->l && e->f < e->c) {
                e->t->l = v->l + 1;
                if (e->t == t) return true;
                else q.push(e->t);
            }
        }

        return false;
    }

    int findPath(Node *s, Node *t, const int limit = INT_MAX) {
        if (s == t) return limit;

        for (Edge *&e = s->c; e; e = e->next) if (e->t->l == s->l + 1 && e->f < e->c) {
            int f = findPath(e->t, t, std::min(limit, e->c - e->f));
            if (f) {
                e->f += f, e->r->f -= f;
                return f;
            }
        }

        return 0;
    }

    int operator()(const int s, const int t, const int n) {
        int res = 0;
        while (makeLevelGraph(&N[s], &N[t], n)) {
            int f;
            while ((f = findPath(&N[s], &N[t])) > 0) res += f;
        }

        return res;
    }
} dinic;

inline void addEdge(const int s, const int t, const int c) {
    N[s].e = new Edge(&N[s], &N[t], c);
    N[t].e = new Edge(&N[t], &N[s], 0);
    (N[s].e->r = N[t].e)->r = N[s].e;
}

int s, t;

int main() {
    int n, m1, m2;
    scanf("%d %d %d", &n, &m1, &m2);

    const int s = 0, t = n * 2 + m1 + m2 + 1;

    for (int i = 1; i <= n; i++) addEdge(i, i + n, 1);
    for (int i = n + n + 1; i <= n + n + m1; i++) addEdge(s, i, 1);
    for (int i = n + n + m1 + 1; i <= n + n + m1 + m2; i++) addEdge(i, t, 1);

    for (int i = 1; i <= n; i++) {
        int k1, k2;
        scanf("%d %d", &k1, &k2);

        for (int j = 0; j < k1; j++) {
            int x;
            scanf("%d", &x);
            addEdge(n + n + x, i, 1);
        }

        for (int j = 0; j < k2; j++) {
            int x;
            scanf("%d", &x);
            addEdge(i + n, n + n + m1 + x, 1);
        }
    }