自动搬运

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来自洛谷，原作者为

	EuphoricStar Remember.

搬运于2025-08-24 21:13:50，当前版本为作者最后更新于2021-07-05 07:06:59，作者可能在搬运后再次修改，您可在原文处查看最新版

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以下是正文

思路

本题可转化为最大流问题。

设源点为 $1$，汇点为 $nl + nr + 2$，左侧点编号为 $2$ ~ $nl + 1$，右侧点编号为 $nl + 2$ ~ $nl + nr + 1$。

然后源点向每个左侧点连一条容量为 $1$ 的边，每个右侧点向汇点连一条容量为 $1$ 的边，左侧点和右侧点按照输入的边连一条容量为 $1$ 的边，然后跑一遍 Dinic，此时最大流便是答案。

代码

#include <bits/stdc++.h>
using namespace std;

const int maxn = 1020, maxm = 511000, inf = 0x3f3f3f3f;

int n, nl, nr, m, s, t, head[maxn], len;
struct edge {
    int from, to, next, cap, flow;
} edges[maxm];

void add_edge(int u, int v, int c, int f) {
    edges[++len].from = u;
    edges[len].to = v;
    edges[len].next = head[u];
    edges[len].cap = c;
    edges[len].flow = f;
    head[u] = len;
}

struct Dinic {
    bool vis[maxn];
    int d[maxn], cur[maxn];

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

    bool bfs() {
        memset(vis, 0, sizeof(vis));
        queue<int> q;
        d[s] = 0;
        vis[s] = 1;
        q.push(s);
        while (q.size()) {
            int u = q.front();
            q.pop();
            for (int i = head[u]; i; i = edges[i].next) {
                edge e = edges[i];
                if (!vis[e.to] && e.cap > e.flow) {
                    vis[e.to] = 1;
                    d[e.to] = d[u] + 1;
                    q.push(e.to);
                }
            }
        }
        return vis[t];
    }

    int dfs(int u, int a) {
        if (u == t || !a) {
            return a;
        }
        int flow = 0, f;
        for (int &i = cur[u]; i; i = edges[i].next) {
            edge &e = edges[i];
            if (d[u] + 1 == d[e.to] && (f = dfs(e.to, min(a, e.cap - e.flow))) > 0) {
                e.flow += f;
                edges[i ^ 1].flow -= f;
                flow += f;
                a -= f;
                if (!a) {
                    break;
                }
            }
        }
        return flow;
    }

    int solve() {
        int flow = 0;
        while (bfs()) {
            for (int i = 1; i <= n; ++i) {
                cur[i] = head[i];
            }
            flow += dfs(s, inf);
        }
        return flow;
    }
} solver;

int main() {
    scanf("%d%d%d", &nl, &nr, &m);
    s = 1;
    t = n = nl + nr + 2;
    len = 1;
    for (int i = 1; i <= nl; ++i) {
        solver.add(s, i + 1, 1);
    }
    for (int i = 1; i <= nr; ++i) {
        solver.add(i + nl + 1, t, 1);
    }
    while (m--) {
        int u, v;
        scanf("%d%d", &u, &v);
        solver.add(u + 1, v + nl + 1, 1);
    }
    printf("%d", solver.solve());
    return 0;
}