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「ZJOI2010」网络扩容 - 网络流 + 费用流

给定一张有向图,每条边都有一个容量 C C 和一个扩容费用 W W 。这里扩容费用是指将容量扩大 1 1 所需的费用,求

  1. 在不扩容的情况下,1 1 N N 的最大流;
  2. 1 1 N N 的最大流增加 K K 所需的最小扩容费用。

链接

BZOJ 1834

题解

第一问直接跑网络流。

第二问,考虑扩容的实质是增加一条连接原来两个点的边,并且这条边费用为 W W
将每一条边对应新增的边加入到残量网络中,限制流量为 K K 跑最小费用流即可。

代码

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

const int MAXN = 1000;
const int MAXM = 5000;
const int MAXK = 10;

struct Node;
struct Edge;

struct Node {
    Edge *e, *c, *in;
    int l, d, f;
    bool q;
} N[MAXN + 1];

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

    Edge(Node *s, Node *t, const int c, const int w) : s(s), t(t), f(0), c(c), w(w), 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 edmondskarp(const int s, const int t, const int n, int &f, int &c) {
    f = c = 0;
    while (true) {
        for (int i = 0; i < n; i++) {
            N[i].f = 0;
            N[i].d = INT_MAX;
            N[i].q = false;
            N[i].in = NULL;
        }

        std::queue<Node *> q;

        q.push(&N[s]);

        N[s].d = 0, N[s].f = INT_MAX;

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

            v->q = false;

            for (Edge *e = v->e; e; e = e->next) if (e->f < e->c && e->t->d > v->d + e->w) {
                e->t->d = v->d + e->w;
                e->t->in = e;
                e->t->f = std::min(v->f, e->c - e->f);
                if (!e->t->q) {
                    e->t->q = true;
                    q.push(e->t);
                }
            }
        }

        if (N[t].d == INT_MAX) return;

        for (Edge *e = N[t].in; e; e = e->s->in) {
            e->f += N[t].f;
            e->r->f -= N[t].f;
        }

        f += N[t].f;
        c += N[t].f * N[t].d;
    }
}

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

int main() {
    int n, m, k;
    static struct Edge {
        int s, t, c, w;
    } E[MAXM];

    scanf("%d %d %d", &n, &m, &k);
    for (int i = 0; i < m; i++) {
        Edge &e = E[i];
        scanf("%d %d %d %d", &e.s, &e.t, &e.c, &e.w);
        addEdge(e.s, e.t, e.c, 0);
    }

    int maxFlow = dinic(1, n, n + 1);

    for (int i = 0; i < m; i++) {
        Edge &e = E[i];
        addEdge(e.s, e.t, INT_MAX, e.w);
    }

    addEdge(0, 1, k, 0);

    int flow, cost;
    edmondskarp(0, n, n + 1, flow, cost);
    printf("%d %d\n", maxFlow, cost);

    return 0;
}